Organic Chemistry
With a Biological Emphasis
Volume II: Chapters 9-17
Tim Soderberg
University of Minnesota, Morris
January 2016
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Notes to the reader:
This textbook is intended for a sophomore-level, two-semester course in Organic
Chemistry targeted at Biology, Biochemistry, and Health Science majors. It is assumed
that readers have taken a year of General Chemistry and college level Introductory
Biology, and are concurrently enrolled in the typical Biology curriculum for sophomore
Biology/Health Sciences majors.
This textbook is meant to be a constantly evolving work in progress, and as such,
feedback from students, instructors, and all other readers is greatly appreciated. Please
send any comments, suggestions, or notification of errors to the author at
soderbt@morris.umn.edu.
If you are looking at a black and white printed version of this textbook, please be aware
that most of the figures throughout are meant to contain color, which is used to help the
reader to understand the concepts being illustrated. It will often be very helpful to refer to
the full-color figures in a digital version of the book, either at the Chemwiki site (see
below) or in a PDF version which is available for free download at:
http://facultypages.morris.umn.edu/~soderbt/textbook_website.htm
An online version is accessible as part of the Chemwiki project at the University of
California, Davis:
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Bi
ological_Emphasis.
This online version contains some additional hyperlinks to animations, interactive 3D
figures, and online lectures that you may find useful. Note: The online (Chemwiki)
version currently corresponds to the older (2012) edition of this textbook. It is scheduled
to be updated to this 2016 edition during the spring and summer of 2016.
Where is the index? There is no printed index. However, an electronic index is available
simply by opening the digital (pdf) version of the text (see above) and using the 'find' or
'search' function of your pdf viewer.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
Volume I: Chapters 1-8
Chapter 1: Introduction to organic structure and bonding, part I
Introduction: Pain, pleasure, and organic chemistry: the sensory effects of capsaicin and vanillin
Section 1: Drawing organic structures
A: Formal charge
B: Common bonding patterns in organic structures
C: Using the 'line structure' convention
D: Constitutional isomers
Section 2: Functional groups and organic nomenclature
A: Functional groups in organic compounds
B: Naming organic compounds
C: Abbreviating organic structure drawings
Section 3: Structures of some important classes of biological molecules
A: Lipids
B: Biopolymer basics
C: Carbohydrates
D: Amino acids and proteins
E: Nucleic acids (DNA and RNA)
Chapter 2: Introduction to organic structure and bonding, part II
Introduction: Moby Dick, train engines, and skin cream
Section 1: Covalent bonding in organic molecules
A: The σ bond in the H2 molecule
B: sp3 hybrid orbitals and tetrahedral bonding
C: sp2 and sp hybrid orbitals and π bonds
Section 2: Molecular orbital theory
A: Another look at the H2 molecule using molecular orbital theory
B: MO theory and conjugated π bonds
C: Aromaticity
Section 3: Resonance
A: What is resonance?
B: Resonance contributors for the carboxylate group
C: Rules for drawing resonance structures
D: Major vs minor resonance contributors
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
Section 4: Non-covalent interactions
A: Dipoles
B: Ion-ion, dipole-dipole and ion-dipole interactions
C: Van der Waals forces
D: Hydrogen bonds
E: Noncovalent interactions and protein structure
Section 5: Physical properties of organic compounds
A: Solubility
B: Boiling point and melting point
C: Physical properties of lipids and proteins
Chapter 3: Conformation and Stereochemistry
Introduction: Louis Pasteur and the discovery of molecular chirality
Section 1: Conformations of open-chain organic molecules
Section 2: Conformations of cyclic organic molecules
Section 3: Chirality and stereoisomers
Section 4: Labeling chiral centers
Section 5: Optical activity
Section 6: Compounds with multiple chiral centers
Section 7: Meso compounds
Section 8: Fischer and Haworth projections
Section 9: Stereochemistry of alkenes
Section 10: Stereochemistry in biology and medicine
Section 11: Prochirality
A: pro-R and pro-S groups on prochiral carbons
B: The re and si faces of carbonyl and imine groups
Chapter 4: Structure determination part I - Infrared spectroscopy, UV-visible spectroscopy,
and mass spectrometry
Introduction: A foiled forgery
Section 1: Mass Spectrometry
A: An overview of mass spectrometry
B: Looking at mass spectra
C: Gas chromatography-mass spectrometry
D: Mass spectrometry of proteins - applications in proteomics
Section 2: Introduction to molecular spectroscopy
A: The electromagnetic spectrum
B: Overview of the molecular spectroscopy experiment
Section 3: Infrared spectroscopy
Section 4: Ultraviolet and visible spectroscopy
A: The electronic transition and absorbance of light
B: Looking at UV-vis spectra
C: Applications of UV spectroscopy in organic and biological chemistry
ii
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
Chapter 5: Structure determination part II - Nuclear magnetic resonance spectroscopy
Introduction: Saved by a sore back
Section 1: The origin of the NMR signal
A: The magnetic moment
B: Spin states and the magnetic transition
Section 2: Chemical equivalence
Section 3: The 1H-NMR experiment
Section 4: The basis for differences in chemical shift
A: Diamagnetic shielding and deshielding
B: Diamagnetic anisotropy
C: Hydrogen-bonded protons
Section 5: Spin-spin coupling
Section 6: 13C-NMR spectroscopy
Section 7: Solving unknown structures
Section 8: Complex coupling in 1H-NMR spectra
Section 9: Other applications of NMR
A: Magnetic Resonance Imaging
B: NMR of proteins and peptides
Chapter 6: Overview of organic reactivity
Introduction: The $300 million reaction
Section 1: A first look at some organic reaction mechanisms
A: The acid-base reaction
B: A one-step nucleophilic substitution mechanism
C: A two-step nucleophilic substitution mechanism
Section 2: A quick review of thermodynamics and kinetics
A: Thermodynamics
B: Kinetics
Section 3: Catalysis
Section 4: Comparing biological reactions to laboratory reactions
Chapter 7: Acid-base reactions
Introduction: A foul brew that shed light on an age-old disease
Section 1: Acid-base reactions
A: The Brønsted-Lowry definition of acidity
B: The Lewis definition of acidity
Section 2: Comparing the acidity and basicity of organic functional groups– the acidity constant
A: Defining Ka and pKa
B: Using pKa values to predict reaction equilibria
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
iii
Table of Contents
C: Organic molecules in buffered solution: the Henderson-Hasselbalch equation
Section 3: Structural effects on acidity and basicity
A: Periodic trends
B: Resonance effects
C: Inductive effects
Section 4: Acid-base properties of phenols
Section 5: Acid-base properties of nitrogen-containing functional groups
A: Anilines
B: Imines
C: Pyrroles
Section 6: Carbon acids
A: The acidity of α-protons
B: Keto-enol tautomers
C: Imine-enamine tautomers
D: The acidity of terminal alkynes
Section 7: Polyprotic acids
Section 8: Effects of enzyme microenvironment on acidity and basicity
Chapter 8: Nucleophilic substitution reactions
Introduction: Why aren't identical twins identical? Just ask SAM.
Section 1: Two mechanistic models for nucleophilic substitution
A: The SN2 mechanism
B: The SN1 mechanism
Section 2: Nucleophiles
A: What is a nucleophile?
B: Protonation state
C: Periodic trends in nucleophilicity
D: Resonance effects on nucleophilicity
E: Steric effects on nucleophilicity
Section 3: Electrophiles
Section 4: Leaving groups
Section: 5: Carbocation stability
Section 6: SN1 reactions with allylic electrophiles
Section 7: SN1 or SN2? Predicting the mechanism
Section 8: Biological nucleophilic substitution reactions
A: A biochemical SN2 reaction
B: A biochemical SN1 reaction
C: A biochemical SN1/SN2 hybrid reaction
Section 9: Nucleophilic substitution in the lab
A: The Williamson ether synthesis
B: Turning a poor leaving group into a good one: tosylates
iv
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
Volume II: Chapters 9-17
Chapter 9: Phosphate transfer reactions
Introduction: Does ET live in a lake in central California?
Section 1: Overview of phosphate groups
A: Terms and abbreviations
B: Acid constants and protonation states
C: Bonding in phosphates
Section 2: Phosphate transfer reactions - an overview
Section 3: ATP, the principal phosphate group donor
Section 4: Phosphorylation of alcohols
Section 5: Phosphorylation of carboxylates
Section 6: Hydrolysis of organic phosphates
Section 7: Phosphate diesters in DNA and RNA
Section 8: The organic chemistry of genetic engineering
Chapter 10: Nucleophilic carbonyl addition reactions
Introduction: How much panda power will your next car have?
Section 1: Nucleophilic additions to aldehydes and ketones: an overview
A: The aldehyde and ketone functional groups
B: Nucleophilic addition
C: Stereochemistry of nucleophilic addition
Section 2: Hemiacetals, hemiketals, and hydrates
A: Overview
B: Sugars as intramolecular hemiacetals and hemiketals
Section 3: Acetals and ketals
A: Overview
B: Glycosidic bond formation
C: Glycosidic bond hydrolysis
Section 4: N-glycosidic bonds
Section 5: Imines
Section 5: A look ahead: addition of carbon and hydride nucleophiles to carbonyls
Chapter 11: Nucleophilic acyl substitution reactions
Introduction: A mold that has saved millions of lives: the discovery of penicillin
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
v
Table of Contents
Section 1: Carboxylic acid derivatives
Section 2: The nucleophilic acyl substitution mechanism
Section 3: The relative reactivity of carboxylic acid derivatives
Section 4: Acyl phosphates
Section 5: Formation of thioesters, esters, and amides
A: Thioester formation
B: Ester formation
C: Amide formation
Section 6: Hydrolysis of thioesters, esters, and amides
Section 7: Protein synthesis on the ribosome
Section 8: Nucleophilic acyl substitution reactions in the laboratory
A: Ester reactions: bananas, soap and biodiesel
B: Acid chlorides and acid anhydrides
C: Synthesis of polyesters and polyamides
D: The Gabriel synthesis of primary amines
Section 9: A look ahead: acyl substitution reactions with a carbanion or hydride ion nucleophile
Chapter 12: Reactions at the alpha carbon, part I
Introduction: A killer platypus and the hunting magic
Section 1: Review of acidity at the α-carbon
Section 2: Isomerization at the α-carbon
A: Carbonyl regioisomerization
B: Stereoisomerization at the α-carbon
C: Alkene regioisomerization
Section 3: Aldol addition
A: Overview of the aldol addition reaction
B: Biochemical aldol addition
C: Going backwards: retroaldol cleavage
D: Aldol addition reactions with enzyme-linked enamine intermediates
Section 4: α-carbon reactions in the synthesis lab - kinetic vs. thermodynamic alkylation products
Interchapter: Predicting multistep pathways - the retrosynthesis approach
Chapter 13: Reactions at the α-carbon, part II
Introduction: The chemistry behind Lorenzo's Oil
Section 1: Decarboxylation
Section 2: An overview of fatty acid metabolism
Section 3: Claisen condensation
A: Claisen condensation - an overview
B: Biochemical Claisen condensation examples
vi
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
C: Retro-Claisen cleavage
Section 4: Conjugate addition and elimination
Section 5: Carboxylation
A: Rubisco, the 'carbon fixing' enzyme
B: Biotin-dependent carboxylation
Chapter 14: Electrophilic reactions
Introduction: Satan Loosed in Salem
Section 1: Electrophilic addition to alkenes
A: Addition of HBr
B: The stereochemistry of electrophilic addition
C: The regiochemistry of electrophilic addition
D: Addition of water and alcohol
E: Addition to conjugated alkenes
F: Biochemical electrophilic addition reactions
Section 2: Elimination by the E1 mechanism
A: E1 elimination - an overview
B: Regiochemistry of E1 elimination
C: Stereochemistry of E1 elimination
D: The E2 elimination mechanism
E: Competition between elimination and substitution
F: Biochemical E1 elimination reactions
Section 3: Electrophilic isomerization
Section 4: Electrophilic substitution
A: Electrophilic substitution reactions in isoprenoid biosynthesis
B: Electrophilic aromatic substitution
Section 5: Carbocation rearrangements
Chapter 15: Oxidation and reduction reactions
Introduction: How to give a mouse a concussion
Section 1: Oxidation and reduction of organic compounds - an overview
Section 2: Oxidation and reduction in the context of metabolism
Section 3: Hydrogenation of carbonyl and imine groups
A: Overview of hydrogenation and dehydrogenation
B: Nicotinamide adenine dinucleotide - a hydride transfer coenzyme
C: Stereochemistry of ketone hydrogenation
D: Examples of biochemical carbonyl/imine hydrogenation
E: Reduction of ketones and aldehydes in the laboratory
Section 4: Hydrogenation of alkenes and dehydrogenation of alkanes
A: Alkene hydrogenation
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
vii
Table of Contents
B: Flavin-dependent alkane dehydrogenation
Section 5: Monitoring hydrogenation and dehydrogenation reactions by UV spectroscopy
Section 6: Redox reactions of thiols and disulfides
Section 7: Flavin-dependent monooxygenase reactions: hydroxylation, epoxidation, and the
Baeyer-Villiger oxidation
Section 8: Hydrogen peroxide is a harmful 'Reactive Oxygen Species'
Chapter 16: Radical reactions
Introduction: The scourge of the high seas
Section 1: Overview of single-electron reactions and free radicals
Section 2: Radical chain reactions
Section 3: Useful polymers formed by radical chain reactions
Section 4: Destruction of the ozone layer by a radical chain reaction
Section 5: Oxidative damage to cells, vitamin C, and scurvy
Section 6: Flavin as a one-electron carrier
Chapter 17: The organic chemistry of vitamins
Introduction: The Dutch Hunger Winter and prenatal vitamin supplements
Section 1: Pyridoxal phosphate (Vitamin B6)
A: PLP in the active site: the imine linkage
B: PLP-dependent amino acid racemization
C: PLP-dependent decarboxylation
D: PLP-dependent retroaldol and retro-Claisen cleavage
E: PLP-dependent transamination
F: PLP-dependent β-elimination and β-substitution
G: PLP-dependent γ-elimination and γ-substitution reactions
H: Racemase to aldolase: altering the course of a PLP reaction
I: Stereoelectronic considerations of PLP-dependent reactions
Section 2: Thiamine diphosphate (Vitamin B1)
Section 3: Thiamine diphosphate, lipoamide and the pyruvate dehydrogenase reaction
Section 4: Folate
A: Active forms of folate
B: Formation of formyl-THF and methylene-THF
C: Single-carbon transfer with formyl-THF
D: Single-carbon transfer with methylene-THF
viii
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Table of Contents
Appendix: Enzymatic reactions by metabolic pathway and EC number
Tables
Table 1: Some characteristic absorption frequencies in IR spectroscopy
Table 2: Typical values for 1H-NMR chemical shifts
Table 3: Typical values for 13C-NMR chemical shifts
Table 4: Typical coupling constants in NMR
Table 5: The 20 common amino acids
Table 6: Structures of common coenzymes
Table 7: Representative acid constants
Table 8: Some common laboratory solvents, acids, and bases
Table 9: Functional groups in organic chemistry
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
ix
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
Chapter 9
Phosphate transfer reactions
Introduction
This chapter is about the chemistry of phosphates, a ubiquitous functional group in
biomolecules that is based on phosphoric acid:
O
HO P OH
OH
phosphoric acid
fig 1d
In late 2010, people around the world found themselves getting a crash course in
phosphate chemistry as they watched the evening news. Those who paid close attention
to the developing story also got an interesting glimpse into the world of scientific
research and debate.
It all started when the American National Aeronautics and Space Administration (NASA)
released the following statement to the news media:
“NASA will hold a news conference at 2 p.m. EST on Thursday, Dec. 2, to
discuss an astrobiology finding that will impact the search for evidence of
extraterrestrial life.”
The wording of the statement attracted widespread media attention, and had some people
holding their breath in anticipation that NASA would be introducing a newly discovered
alien life form to the world. When December 2nd came, however, those hoping to meet
ET were disappointed – the life form being introduced was a bacterium, and it was from
our own planet. To biologists and chemists, though, the announcement was nothing less
than astounding.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
1
Chapter 9: Phosphate transfer
The NASA scientists worked hard to emphasize the significance of their discovery during
the news conference. Dr. Felicia Wolfe-Simon, a young postdoctoral researcher who had
spearheaded the project, stated that they had “cracked open the door to what's possible for
life elsewhere in the universe - and that's profound". A senior NASA scientist claimed
that their results would "fundamentally change how we define life", and, in attempting to
convey the importance of the discovery to a reporter from the newspaper USA Today,
referred to an episode from the original Star Trek television series in which the crew of
the Starship Enterprise encounters a race of beings whose biochemistry is based on silica
rather than carbon.
The new strain of bacteria, dubbed 'GFAJ-1', had been isolated from the arsenic-rich mud
surrounding salty, alkaline Mono Lake in central California. What made the strain so
unique, according to the NASA team, was that it had evolved the ability to substitute
arsenate for phosphate in its DNA. Students of biology and chemistry know that
phosphorus is one of the six elements that are absolutely required for life as we know it,
and that DNA is a polymer linked by phosphate groups. Arsenic, which is directly below
phosphorus on the periodic table, is able to assume a bonding arrangement like that of
phosphate, so it might seem reasonable to wonder whether arsenate could replace
phosphate in DNA and other biological molecules. Actually finding a living thing with
arsenate-linked DNA would indeed be a momentous achievement in biology, as this
would represent a whole new chemistry for the most fundamental molecule of life, and
would change our understanding of the chemical requirements for life to exist on earth and potentially other planets.
In 1987, Professor F.H. Westheimer of Harvard University published what would become
a widely read commentary in Science Magazine entitled “Why Nature Chose
Phosphates”. In it, he discussed the chemical properties that make the phosphate group so
ideal for the many roles that it plays in biochemistry, chief among them the role of a
linker group for DNA polymers. One of the critical characteristics of phosphate that
Westheimer pointed out was that the bonds linking phosphate to organic molecules are
stable in water. Clearly, if you are selecting a functional group to link your DNA, you
don't want to choose one that will rapidly break apart in water. Among the functional
groups that Westheimer compared to phosphate in terms of its suitability as a potential
DNA linker was arsenate –but he very quickly dismissed the idea of arsenate-linked
DNA because it would be far too unstable in water.
Given this background, it is not hard to imagine that many scientists were puzzled, to say
the least, by the NASA results. While the popular media took the announcement at face
value and excitedly reported the results as a monumental discovery – NASA is, after all,
a highly respected scientific body and the study was being published in Science
Magazine, one of the most prestigious scientific journals in the world – many scientists
quickly voiced their skepticism, mainly in the relatively new and unconstrained venue of
the blogosphere. Microbiologist Rosie Redfield of the University of British Columbia,
writing in her blog devoted to 'open science', wrote a detailed and highly critical analysis
of the study. She pointed out, among other things, that the experimenters had failed to
2
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
perform the critical purification and mass spectrometry analyses needed to demonstrate
that arsenate was indeed being incorporated into the DNA backbone, and that the broth in
which the bacteria were being grown actually contained enough phosphate for them to
live and replicate using normal phosphate-linked DNA. Science journalist Carl Zimmer,
in a column in the online magazine Slate, contacted twelve experts to get their opinions,
and they were overwhelmingly negative. One of the experts said bluntly, “This paper
should not have been published". Basically, the NASA researchers were making an
astounding claim that, if true, would refute decades of established knowledge about the
chemistry of DNA – but the evidence they presented was far from convincing. Carl
Sagan's widely quoted dictum - “extraordinary claims require extraordinary evidence” seemed to apply remarkably well to the situation.
What followed was a very public, very lively, and not always completely collegial debate
among scientists about the proper way to discuss science: the NASA researchers
appeared to dismiss the criticism amassed against their study because it came from blogs,
websites, and Twitter feeds. The proper venue for such discussion, they claimed, was in
the peer-reviewed literature. Critics countered that their refusal to respond to anything
outside of the traditional peer-review system was disingenuous, because they had made
full use of the publicity-generating power of the internet and mainstream media in the
first place when they announced their results with such fanfare.
The traditional venue for debate, while quite a bit slower than the blogosphere, did
eventually come through. When the full paper was published in Science a few months
later, it was accompanied by eight 'technical comments' from other researchers pointing
out deficiencies in the study, an 'editors note', and a broader news article about the
controversy. In July of 2012, a paper was published in Science under the title “GFAJ-1 Is
an Arsenate-Resistant, Phosphate-Dependent Organism”. The paper reported definitive
evidence that DNA from GFAJ-1, under the conditions described in the NASA paper, did
not have arsenate incorporated into its structure. Just like professor Westheimer discussed
in the 1980s, it appears that nature really did choose phosphate – and only phosphate –
after all . . . at least on this planet.
Background reading and viewing:
Youtube video of the NASA press conference:
http://www.youtube.com/watch?v=WVuhBt03z8g.
Wolfe-Simon, F. et al. Science Express, Dec 2, 2010. The first preview article on the
proposed 'arsenic bacteria'.
Wolfe-Simon, F. et al., Science 2011, 332, 1163. The full research paper in Science
Magazine.
Westheimer, F.H. Science 1987, 235, 1173. The article by Westheimer titled 'Why
Nature Chose Phosphates'.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
3
Chapter 9: Phosphate transfer
Zimmer, Carl, Slate, Dec 7, 2010: Blog post by Carl Zimmer titled 'This Paper Should
Not Have Been Published'.
http://www.slate.com/articles/health_and_science/science/2010/12/this_paper_should_no
t_have_been_published.html
Redfield, R. Blog post Dec 4, 2010:
http://rrresearch.fieldofscience.com/2010/12/arsenic-associated-bacteria-nasas.html
Science 2012, 337, 467. The paper in Science Magazine refuting the validity of the
arsenic bacteria claim.
Section 9.1: Overview of phosphate groups
Phosphate is everywhere in biochemistry. As we were reminded in the introduction to
this chapter, our DNA is linked by phosphate:
H
DNA
N
O
O
O
O
N
O
CH3
N
O
O P O
O
O
NH2
N
O
DNA
fig 1a
The function of many proteins is regulated - switched on and off - by enzymes which
attach or remove a phosphate group from the side chains of serine, threonine, or tyrosine
residues.
4
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
O
OH
P
O
O O
protein
protein
protein
H
protein
N
N
H
O
O
phosphotyrosine residue
tyrosine residue
fig 1b
Countless diseases are caused by defects in phosphate transferring enzymes. As just one
example, achondroplasia, a common cause of dwarfism, is caused by a defect in an
enzyme whose function is to transfer a phosphate to a tyrosine residue in a growth-related
signaling protein.
Recall from section 8.4 that phosphates are excellent leaving groups in biological organic
reactions, as the negative charges can be effectively stabilized through interactions in an
enzyme active site with bound magnesium ion and amino acid groups.
Clearly, an understanding of phosphate chemistry is central to the study of biological
organic chemistry. We'll begin with an overview of terms used when talking about
phosphates.
9.1A: Terms and abbreviations
The fully deprotonated conjugate base of phosphoric acid is called a phosphate ion, or
inorganic phosphate (often abbreviated 'Pi'). When two phosphate groups are linked to
each other, the linkage itself is referred to as a 'phosphate anhydride', and the
compound is called 'inorganic pyrophosphate' (often abbreviated PPi).
phosphate anhydride
linkage
O
O
HO P OH
O P O
OH
O
phosphoric acid
inorganic phosphate (Pi)
O
O
O P O P O
O
O
inorganic pyrophosphate (PPi)
fig 1
The chemical linkage between phosphate and a carbon atom is a phosphate ester.
Adenosine monophosphate (AMP) has a single phosphate ester linkage.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
5
Chapter 9: Phosphate transfer
phosphate ester
NH2
N
N
O
O P O
N
O
N
O
HO
OH
adenosine monophosphate (AMP)
fig 2
Adenosine triphosphate has one phosphate ester linkage and two phosphate anhydride
linkages.
phosphate anhydrides
O
O
phosphate ester
N
N
O
O P O P O P O
O
NH2
O
O
N
N
O
HO
OH
adenosine triphosphate (ATP)
fig 3
Oxygen atoms in phosphate groups are referred to either as 'bridging' or 'non-bridging',
depending on their position. An organic diphosphate has two bridging oxygens (one in
the phosphate ester linkage and one in the phosphate anhydride linkage) and five nonbridging oxygens:
O
O
R O P O P O
O
red = bridging oxygen
blue = non-bridging oxygen
O
fig 4
A single phosphate is linked to two organic groups is called phosphate diester. The
backbone of DNA is linked by phosphate diesters.
6
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
DNA
O
Base
O
O
OR1
O P OR2
=
O P O
O
O
O
Base
phosphate diester
HO
fig 5
Organic phosphates are often abbreviated using 'OP' and 'OPP' for mono- and
diphosphates, respectively. For example, glucose-6-phosphate and isopentenyl
diphosphate are often depicted as shown below. Notice that the 'P' abbreviation includes
the associated oxygen atoms and negative charges.
O
O P O
OP
O
O
HO
OH
HO
=
O
HO
OH
HO
OH
OH
glucose-6-phosphate
O
=
PPO
isopentenyl diphosphate
O
O P O P O
O
O
fig 6
Exercise 9.1: Consider the biological compounds below, some of which are shown with
abbreviated structures:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
7
Chapter 9: Phosphate transfer
O
O
O P O
PO
O
O
OH
HO
OH
PPPO
O
H3N
OP
O
O
O
OH
OH
b)
a)
N H
N
c)
NH2
N
O HO CH3
O
HO
O
N
N
N
NH2
OPP
N
d)
O
N
OH
O P O
O
N
N
O
HO
OH
e)
fig 4a
a) Which contain one or more phosphate anhydride linkages? Specify the number of
phosphate anhydride linkages in your answers.
b) Which contain one or more phosphate monoesters? Again, specify the number for each
answer.
c) Which contain a phosphate diester?
d) Which could be described as an organic diphosphate?
e) For each compound, specify the number of bridging and non-bridging oxygens in the
phosphate group.
9.1B: Acid constants and protonation states
Phosphoric acid is triprotic, meaning that it has three acidic protons available to donate,
with pKa values of 1.0, 6.5, and 13.0, respectively. (da Silva and Williams)
8
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
O
O
O
O
HO P OH
O P OH
O P OH
O P O
OH
OH
O
O
H3PO4
pKa = 1.0
H2PO4-1
pKa = 6.5
HPO4-2
pKa = 13.0
PO4-3
fig 7
These acid constant values, along with the Henderson-Hasselbalch equation (section
7.2C) tell us that, at the physiological pH of approximately 7, somewhat more than half
of the phosphate species will be in the HPO4-2 state, and slightly less than half will be in
the H2PO4-1 state, meaning that the average net charge is between -1.5 and -2.0.
Phosphate diesters have a pKa of about 1, meaning that they carry a full negative charge
at physiological pH.
OR
OR
O P OR
O P OR
OH
O
pKa ~ 1
deprotonated at pH 7
fig 7a
Organic monophosphates, diphosphates, and triphosphates all have net negative charges
and are partially protonated at physiological pH, but by convention are usually drawn in
the fully deprotonated state.
Exercise 9.2: Explain why the second pKa of phosphoric acid is so much higher than the
first pKa.
Exercise 9.3: What is the approximate average charge of inorganic phosphate in a
solution buffered to pH 2?
9.1C: Bonding in phosphates
Looking at the location of phosphorus on the periodic table, you might expect it to bond
and react in a fashion similar to nitrogen, which is located just above it in the same
column. Indeed, phosphines - phosphorus analogs of amines - are commonly used in the
organic laboratory.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
9
Chapter 9: Phosphate transfer
H3C N CH
3
H3C
H3C P CH
3
H3C
trimethylamine
trimethylphosphine
fig 8
However it is in the form of phosphate, rather than phosphine, that phosphorus plays its
main role in biology.
The four oxygen substituents in phosphate groups are arranged about the central
phosphorus atom with tetrahedral geometry, however there are a total of five bonds to
phosphorus - four σ bonds and one delocalized π bond.
O
O
O P O
O
O P O
O
O
O P O
O
O
O
O P O
O
O P O
O
-3
O
=
O
P
O
O
fig 9
Phosphorus can break the 'octet rule' because it is on the third row of the periodic table,
and thus has d orbitals available for bonding. The minus 3 charge on a fully deprotonated
phosphate ion is spread evenly over the four oxygen atoms, and each phosphorus-oxygen
bond can be considered to have 25% double bond character: in other words, the bond
order is 1.25.
Recall from section 2.1 the hybrid bonding picture for the tetrahedral nitrogen in an
amine group: a single 2s and three 2p orbitals combine to form four sp3 hybrid orbitals,
three of which form σ bonds and one of which holds a lone pair of electrons.
N
Nitrogen:
lone pair
2s
2px
2py
2pz
hybridize to sp3
fig 10
10
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
sp3
Chapter 9: Phosphate transfer
In the hybrid orbital picture for phosphate ion, a single 3s and three 3p orbitals also
combine to form four sp3 hybrid orbitals with tetrahedral geometry. In contrast to an
amine, however, four of the five valance electrons on phosphorus occupy sp3 orbitals, and
the fifth occupies an unhybridized 3d orbital.
P
Phosphorus:
3s
3px
3py
3pz
sp3
3d
3d
hybridize to sp3
fig 11
This orbital arrangement allows for four σ bonds with tetrahedral geometry in addition to
a fifth, delocalized π bond formed by π overlap between the half-filled 3d orbital on
phosphorus and 2p orbitals on the oxygen atoms.
In phosphate esters, diesters, and anhydrides the π bonding is delocalized primarily over
the non-bridging bonds, while the bridging bonds have mainly single-bond character. In
a phosphate diester, for example, the two non-bridging oxygens share a -1 charge, as
illustrated by the two major resonance contributors below. The bonding order for the
bridging P-O bonds in a phosphate diester group is about 1, and for the non-bridging P-O
bonds about 1.5. In the resonance contributors in which the bridging oxygens are shown
as double bonds (to the right in the figure below), there is an additional separation of
charge - thus these contributors are minor and make a relatively unimportant contribution
to the overall bonding picture.
O
O
O
O
RO P O
RO
RO P O
RO
RO P O
RO
RO P O
RO
major resonance contributors
O
P
RO
O
RO
minor resonance contributors
(separation of charge)
non-bridging bonds: significant double
bond character
bridging bonds: little double-bond
character
fig 12
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
11
Chapter 9: Phosphate transfer
Exercise 9.4: Draw all of the resonance structures showing the delocalization of charge
on a (fully deprotonated) organic monophosphate. If a 'bond order' of 1.0 is a single bond,
and a bond order of 2.0 is a double bond, what is the approximate bond order of bridging
and non-bridging P-O bonds?
Throughout this book, phosphate groups will often be drawn without attempting to show
tetrahedral geometry, and π bonds and negative charges will usually be shown localized
to a single oxygen. This is done for the sake of simplification - however it is important
always to remember that the phosphate group is really tetrahedral, the negative charges
are delocalized over the non-bridging oxygens, and that there is some degree of
protonation at physiological pH (with the exception of the phosphate diester group).
Section 9.2: Phosphate transfer reactions - an overview
In a phosphate transfer reaction, a phosphate group is transferred from a phosphate
group donor molecule to a phosphate group acceptor molecule:
O
O
OH
P O
O O
O P
O O
phosphate
acceptor
phosphate
donor
+
+
HO
fig 13
A very important aspect of biological phosphate transfer reactions is that the
electrophilicity of the phosphorus atom is usually enhanced by the Lewis acid (electronaccepting) effect of one or more magnesium ions. Phosphate transfer enzymes generally
contain a Mg2+ ion bound in the active site in a position where it can interact with nonbridging phosphate oxygens on the substrate. The magnesium ion pulls electron density
away from the phosphorus atom, making it more electrophilic.
= Mg+2
Mg+2 coordination makes
phosphorus more electrophilic
O
P O
O O
phosphate
donor
O
O
fig 17
12
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
P O
O
phosphate
donor
Chapter 9: Phosphate transfer
Without this metal ion interaction, a phosphate is actually a poor electrophile, as the
negatively-charged oxygens shield the phosphorus center from attack by a nucleophile.
Note: For the sake of simplicity and clarity, we will not draw the magnesium ion or other
active site groups interacting with phosphate oxygens in most figures in this chapter - but
it is important to keep in mind that these interactions play an integral role in phosphate
transfer reactions.
In a phosphate transfer reaction, the phosphorus atom is the electrophilic atom. The
nucleophilic acceptor group approaches the electrophilic, tetrahedral phosphorus center
from the backside, opposite the leaving group (step 1 below). This leads to a trigonal
bipyramidal intermediate, which collapses back to tetrahedral geometery as the leaving
group leaves. Notice the overall similarity to an SN2 mechanism (section 8.1). In
phosphate transfer as with SN2, the overall reaction results in inversion at the tetrahedral
electrophilic atom.
Phosphate transfer mechanism:
acceptor
O
:B
trigonal bipyramidal
intermediate
H
acceptor
O
O P O
O
donor
step 1
O
O
acceptor
donor
O P O
O
step 2
donor
P O
O
+
O
OO
SN2 mechanism:
trigonal bipyramidal transition state
H
HO
H C
H
Cl
H
δδHO C
Cl
H H
H
C
HO
H +
H
Cl
fig 18
fig 19a
In the trigonal bipyramidal intermediate of the phosphate transfer reaction, just as in the
transition state of an SN2 reaction, the five substituents are not equivalent: the three nonbridging oxygens are equatorial (forming the base of the trigonal bipyramid), while the
nucleophile and the leaving group are apical (each forms the apex, or tip, of a pyramid).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
13
Chapter 9: Phosphate transfer
intermediate of phosphate transfer reaction:
equatorial
acceptor
donor
O
O P O
apical
apical
OO
equatorial
fig 19
Exercise 9.5: Predict the approximate angles between the two bonds indicated in a
phosphate transfer intermediate species. Oa refers to an oxygen at the apical position, and
Oe to an oxygen in the equatorial position.
a) Oa-P-Oa
b) Oa-P-Oe
c)Oe-P-Oe
Section 9.3: ATP, the principal phosphate donor
Thus far we have been very general in our discussion of phosphate transfer reactions,
referring only to generic 'donor' and 'acceptor' species. It's time to get more specific. The
most important donor of phosphate groups in the cell is a molecule called adenosine
triphosphate, commonly known by its abbreviation ATP.
triphosphate
ribose
O
O
NH2
N
O
O Pγ O Pβ O Pα O
O
O
O
N
N
O
N
HO
OH
base (adenine)
adenosine triphosphate (ATP)
fig 21
Notice that there are essentially three parts to the ATP molecule: an adenine nucleoside
'base', a five-carbon sugar (ribose), and triphosphate. The three phosphates are designated
by Greek letters α, β, and γ, with the α phosphate being the one closest to the ribose.
14
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
Adenosine diphosphate (ADP) and adenosine monophosphate (AMP) are also important
players in the reactions of this chapter.
ATP is a big molecule, but the bond-breaking and bond-forming events we will be
studying in this chapter all happen in the phosphate part of the molecule. You will see
structural drawings of ATP, ADP, and AMP abbreviated in many different ways in this
text and throughout the biochemical literature, depending on what is being illustrated.
For example, the three structures below are all abbreviated depictions of ATP:
O
O
O
ribose-A
O Pγ O Pβ O Pα O
O
O
O
O
O
O
O Pγ O Pβ O AMP
O Pγ O ADP
O
O
O
fig 22
The following exercise will give you some practice in recognizing different abbreviations
for ATP and other biological molecules that contain phosphate groups.
Exercise 9.6 : Below are a number of representations, labeled A-S, of molecules that
contain phosphate groups. Different abbreviations are used. Arrange A-S into groups of
drawings that depict the same species (for example, group together all of the
abbreviations which depict ATP).
O
R O AMP
O P O ADP
R OPP
OPP
C
D
O
A
B
NH2
N
N
O
R O
P O
O
N
O
N
O
R O P O
F
HO
O
OH
O
O
R O P O P O
O
O
G
E
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
15
Chapter 9: Phosphate transfer
O
O
O
O
O P O P O P O
O
O
ribose-A
R O
O
O
P O
ribose-A
O P
O
H
O AMP
O
J
I
NH2
N
O
O P
O
O P O P
O
N
O
O
O
O
N
O
O
O P
N
O P O
ADP
O
M
O
O
HO
K
L
OH
NH2
N
O
R OP
PPi
N
O
O P O P O
O
N
O
O
N
N
O
HO
OH
P
O
O
O P
O P
O
O
O
O
ribose-A
ATP
O P O P
O
R
Q
O
O AMP
O
S
You are probably familiar with the physiological role of ATP from your biology classes it is commonly called 'the energy currency of the cell'. What this means is that ATP
stores energy we get from the oxidation of fuel molecules such as carbohydrates or fats.
The energy in ATP is stored in the two high-energy phosphate anhydride linkages.
O
O
O
O P O P O P O
O
O
ribose-A
O
the two phosphate anhydride
linkages in ATP
16
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
fig 23
When one or both of these phosphate anhydride links are broken as a phosphate group is
transferred to an acceptor, a substantial amount of energy is released. The negative
charges on the phosphate groups are separated, eliminating some of electrostatic
repulsion that existed in ATP. One way to picture this is as a coil springing open,
releasing potential energy.
phosphate
donor
phosphate
acceptor
O
OH
+
O
O
O Pγ O Pβ O Pα O
O
O
ribose-A
O
ATP
more room for H-bonding to
water in this region after
phosphate transfer
phosphate
acceptor
O
O P O
O
O
O
O Pβ O Pα O
O
ribose-A
O
ADP
repulsing charges are separated
fig 24
In addition, cleavage of a phosphate anhydride bond means that surrounding water
molecules are able to form more stabilizing hydrogen-bonding interactions with the
products than was possible with the starting materials, again making the reaction more
'downhill', or exergonic.
It is important to understand that while the phosphate anhydride bonds in ATP are
thermodynamically unstable (they contain a great deal of chemical energy), they are at
the same time kinetically stable: ATP-cleaving reactions such as the one depicted in
figure 24 (and the DNA-cleaving reactions mentioned in the introduction) are
exothermic, but also have a high energy barrier, making them very slow unless catalyzed
by an enzyme. In other words, the release of the energy contained in ATP is highly
energetic but also subject to tight control by the interaction of highly evolved enzymes in
our metabolic pathways.
ATP is a versatile phosphate group donor: depending on the site of nucleophilic attack (at
the α, β, or γ phosphorus), different phosphate transfer outcomes are possible. Below are
the three most common patterns seen in the central metabolic pathways. A 'squigly' line
in each figure indicates the P-O bond being broken. We will study specific examples of
each of these in the coming sections.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
17
Chapter 9: Phosphate transfer
Attack at the γ-phosphate:
O
acceptor
+
O
O
ribose-A
O Pγ O Pβ O Pα O
OH
O
O
O
Mg+2
attack at Pγ
acceptor
O
O
+
O P O
O
O Pβ O Pα O
O
O
ribose-A
O
ADP
Attack at the β-phosphate:
O
acceptor
+
O
O
OH
O
O
O
Mg+2
acceptor
ribose-A
O Pγ O Pβ O Pα O
O
O
O
O P O P O
O
attack at Pβ
+
O Pα O
ribose-A
O
O
AMP
Attack at the α-phosphate:
18
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
O
acceptor
+
OH
O
O
ribose-A
O Pγ O Pβ O Pα O
O
O
O
Mg+2
attack at Pα
acceptor
O
O
O P O
ribose-A
+
O
O P O P O
O
O
O
PPi
fig 24c
The common thread running through all of the ATP-dependent reactions we will see in
this section is the idea that the phosphate acceptor molecule is undergoing a
thermodynamically 'uphill' transformation to become a more reactive species. The
energy for this uphill transformation comes from breaking a high-energy phosphate
anhydride bond in ATP. That is why ATP is often referred to as 'energy currency': the
energy in its anhydride bonds is used to 'pay for' a thermodynamically uphill chemical
step.
Exercise 9.7: Propose a fourth hypothetical phosphate transfer reaction between ATP and
the generic acceptor molecule in fig 24c, in which inorganic phosphate (Pi) is a byproduct.
Exercise 9.8 : Why is this hypothetical phosphate transfer reaction less energetically
favorable compared to the three possibilities shown in fig 24c??
O
acceptor
OH
+ ATP
O
O
acceptor
O P O P O P O
O
O
+
HO
ribose-A
O
fig 24d
Section 9.4: Phosporylation of alcohols
A broad family of enzymes called kinases catalyze transfer of a phosphate group from
ATP to an alcohol acceptor. Mechanistically, the alcohol oxygen acts as a nucleophile,
attacking the electrophilic γ-phosphorus of ATP and expelling ADP.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
19
Chapter 9: Phosphate transfer
Glucose is phosphorylated in the first step of the glycolysis pathway by the enzyme
hexose kinase (EC 2.7.1.1), forming glucose-6-phosphate.
OH
O
O
HO
HO
O
O Pγ O Pβ O Pα O
O
O
O
+
OH
O
OH
ribose-A
ATP
glucose
attack at Pγ
O
O P O
O
HO
O
O
+
OH
HO
O
O P O P O
α
β
O
O
OH
ribose-A
ADP
glucose 6-phosphate
fig 26c
Here is a shorthand way to depict this reaction. Notice the "ATP in, ADP out" notation
used below, indicating that ATP is one of the reactants and ADP is one of the products.
From here on, we will frequently use this common convention to indicate reaction
participants whose structures are not drawn out in a figure.
one ATP is converted to ADP
OH
ATP
6
HO
O
4
HO
glucose
ADP
O
HO
5
3
OP
OH
2
OH
OH
HO
1
OH
glucose-6-phosphate
fig 26b
The biological activity of many proteins is regulated by protein kinases. In a protein
kinase reaction, the side chain hydroxyl groups on serine, threonine, or tyrosine residues
of certain proteins are phosphorylated by ATP:
20
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
O
ATP
OH
O P O
ADP
O
protein
protein
phosphoserine residue
serine residue
fig 7
fig 27
The conversion of a neutral hydroxyl group to a charged phosphate represents a very
dramatic change in the local architecture of the protein, potentially altering its folding
pattern and ability to bind to small molecules or other proteins. A protein's biological
function can be 'switched on' by phosphorylation of a single residue, and switched off
again by removal of the phosphate group. The latter reaction we will examine later in
this chapter.
Exercise 9.9:
a) Draw a curved-arrow mechanism, using abbreviations as appropriate, for the serine
kinase reaction.
b) Threonine kinase catalyzes the phosphorylation of the side chain hydroxyl group of
threonine residues in proteins. Draw the structure, including the configuration of all
stereocenters, of a phosphothreonine residue. Explain how you can predict the
stereochemistry of the side chain.
Although stereochemical inversion in phosphate transfer is predicted by theory, the fact
that phosphate groups are achiral made it impossible for a long time to verify the
phenomenon directly. This was finally accomplished in the late 1970's, when a group of
researchers demonstrated phosphate inversion in kinase enzymes using chemically
synthesized ATP in which three different isotopes of oxygen were incorporated into the
γ−phosphate, thus creating a chiral phosphorus center. (Ann. Rev. Biochem. 1980 49,
877).
chiral center (R)
(S)
17O
acceptor
OH
+
16O
P
18O
O
ADP
O
acceptor
O
P O
O
+
ADP
labeled ATP
fig 27a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
21
Chapter 9: Phosphate transfer
Alcohols can be converted into organic diphosphates in two different ways. A two-step
process simply involves successive transfers of the γ-phosphate groups of two ATP
donors, such as in these sequential steps in isoprenoid biosynthesis. (EC 2.7.1.36; EC
2.7.4.2). A compound called mevalonate is diphosphorylated in this way in the early
phase of the biosynthetic pathway for cholesterol, steroid hormones, and other isoprenoid
molecules.
ATP
O HO CH3
O
ADP
OH
O HO CH3
O
O P O
O
mevalonate-5-phosphate
mevalonate
O
ATP
ADP
O HO CH3
O
O
O
O P O P O
O
mevalonate-5-diphosphate
O
fig 28
The mechanism for the first phosphorylation step is analogous to that for an alcohol
kinase reaction, which we have just seen. In the second phosphate transfer step,
catalyzed by a separate enzyme, one of the phosphate oxygens on the organic
monophosphate acts as a nucleophilic phosphate acceptor, attacking the γ-phosphate of a
second ATP.
22
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
phosphate oxygen is now the
phosphate acceptor
O HO CH3
O
O
+
O P O
O
mevalonate-5-phosphate
O
O
O P O P O P O
O
O
O
ribose-A
O
ATP
attack at Pγ
O HO
O
O
O
O
O P O P O
O
mevalonate-5-diphosphate
+
O
O P O P O
O
O
ribose-A
O
ADP
fig 29b
In some metabolic pathways, diphosphorylation occurs by a different mechanism from
the one above. Instead of sequentially transferring two phosphates from two ATP
donors, the alternate mechanism occurs in a single step: the nucleophilic acceptor
molecule attacks the β-phosphate of ATP, rather than the γ-phosphate. After formation
of the trigonal bipyramidal intermediate, it is AMP (not ADP) which is expelled, and
what started out as the β and γ phosphates of ATP both remain with the acceptor.
In the biosynthesis of DNA and RNA nucleotides, one of the hydroxyl groups on ribose5-phosphate is diphosphorylated (EC 2.7.6.1) in a one-step mechanism:
O
PO
OH
O
+
HO
O
O
O Pγ O Pβ O Pα O
O
OH
O
ribose-A
O
ribose-5-phosphate
attack at Pβ
O
O
PO
O
O P O P O
O
HO
O
OH
O
O P O
+
ribose-A
O
AMP
phosphoribosyl-5-phosphate
(PRPP)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
23
Chapter 9: Phosphate transfer
fig 31a
The metabolic role of both of the diphosphorylation processes we just saw is to convert a
hydroxyl group into a good leaving group (recall that hydroxide ions are strong bases and
poor leaving groups, while phosphates/diphosphates, when stabilized in an enzyme active
site, are weak bases and very good leaving groups). In nucleoside biosynthesis pathways,
the diphosphate group of PRPP acts as a leaving group in the very next metabolic step,
which is an SN1 reaction: we have already seen this reaction in section 8.8).
Section 9.5: Phosphorylation of carboxylates
Thus far we have seen hydroxyl oxygens and phosphate oxygens acting as nucleophilic
accepting groups in ATP-dependent phosphate transfer reactions. Carboxylate oxygens
can also accept phosphate groups from ATP. This typically happens in two different
ways. First, the carboxylate can attack the γ-phosphate of ATP to accept phosphate,
generates a species known as an 'acyl phosphate'. An example is the first part of the
reaction catalyzed by glutamine synthase (EC 6.3.1.2):
H3N
H
ATP
O
O
CO2
glutamate
ADP
H3N
(attack at Pγ of ATP)
H
O
O
O P O
CO2
O
glutamyl phosphate
Alternatively, carboxylate groups are often converted into a species referred to as an
'acyl-AMP' . Here, the carboxylate oxygen attacks the α-phosphate of ATP leading to
release of inorganic pyrophosphate. An example is the first part of the reaction catalyzed
by the enzyme asparagine synthetase: (EC 6.3.5.4):
24
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
NH3
O
O
O2C
H
aspartate
O Pγ
+
O
O
O
O
Pβ O P O
O
O
ribose-A
α
O
attack at Pα
NH3
O2C
O
O
O
O P O
H
ribose-A
+
O P
O
O
O
P O
O
O
aspartyl-AMP
fig 35
Exercise 9.10: Draw a curved-arrow mechanism for the phosphate transfer reaction
shown below (EC 2.7.2.3), which is from the glycolysis pathway. Note that ADP is on
the reactant side and ATP is a product (the opposite of what we have seen so far). Hint:
What functional group (be specific!) is the nucleophile? What functional group is the
leaving group?
OH
OP
OP
ADP
O
1,3-bisphosphoglycerate
OH
ATP
OH
OP
O
3-phosphoglycerate
fig 36
Section 9.6: Hydrolysis of organic phosphates
While kinase enzymes catalyze the phosphorylation of organic compounds, enzymes
called phosphatases catalyze dephosphorylation reactions. The reactions catalyzed by
kinases and phosphatases are not the reverse of one another: kinases irreversibly
transfer phosphate groups from ATP (or sometimes other nucleoside triphosphates) to
various organic acceptor compounds, while phosphatases transfer phosphate groups from
organic compounds to water: these are hydrolysis reactions. Kinase reactions involve an
inherently 'uphill' step (phosphorylation of an alcohol, for example) being paid for with
an inherently 'downhill' step (cleavage of an anhydride bond in ATP). Phosphatase
reactions, on the other hand, are thermodynamically 'downhill', and while they require an
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
25
Chapter 9: Phosphate transfer
enzyme to speed them up, they do not involve 'spending' energy currency the way kinase
reactions do.
Phosphatase reaction:
O
O
H2O
+
P O
O O R
HO
P O
O
HO
+
R
fig 37
There are two possible general mechanisms for a phosphatase reaction. Some enzymes
catalyze direct hydrolysis reactions, in which the phosphate group is removed by direct
attack of a water molecule at the phosphate center:
Phosphatase mechanism (direct hydrolysis):
B:
H
H
O
O
O
P OR
O O
P
HO
OR
H A
O O
Mg+2
O
O
O
P O
O
HO
P O
O
+
ROH
fig 37
One of the two phosphate groups on fructose 1,6-bisphosphate is hydrolyzed in such a
way late in the gluconeogenesis pathway. (Biochemistry 2000, 39, 8565) (EC 3.1.3.11)
H
H
O
O
+
P
O
HO
OP
O
P O
HO
O
O O
HO
O
+
OH
fructose 1,6-bisphosphate
HO
HO
HO
OP
O
OH
fructose-6-phosphate
fig 37b
Many phosphatase reactions, however, operate by a slightly more complicated
mechanism than what is shown above. In the first phase, a nucleophilic enzyme group
26
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
(typically a cysteine, aspartate, glutamate, or histidine side chain, designated in the figure
below as 'X') attacks the phosphate group. In the second phase, the phosphorylated
residue is hydrolized. For example, protein tyrosine phosphatase catalyzes the
dephosphorylation of phosphotyrosine residues in some proteins - this is the other half of
the regulatory 'on-off switch' that we discussed earlier in the context of protein kinases. In
the first step, the phosphate group is directly donated to a cysteine side chain in the
phosphatase enzyme's active site. In the second step, the phosphocysteine intermediate is
cleaved by water to form inorganic phosphate and regenerate the free cysteine in the
active site.
An indirect phosphatase reaction:
Step 1:
phosphotyrosine on substrate protein
active site cysteine on
phosphatase enzyme
O
enz
O P
+
SH
HO
enz
O
O
S
O
P O
+
O
substrate
protein
substrate
protein
phosphate group has been
transferred to enzyme cysteine
Step 2:
free cysteine is regenerated
enz
S
P O
O
enz
O
+
H2O
SH
O
+
O P OH
O
fig 38d
Notice that in the end, the phosphate group has still been transferred to a water molecule,
albeit indirectly. How would you know, just by looking at the substrate and product of
the protein tyrosine phosphatase reaction, that the phosphate is not transferred directly to
a water molecule? Simply put, you wouldn't know this information without the benefit of
knowledge gained from biochemical experimentation.
Exercise 9.11: If you were to look just at the substrates and products of a phosphatase
reaction without knowing anything about the mechanism, it is apparent that a
nucleophilic substitution mechanism could theoretically account for the products formed.
Draw out a hypothetical nucleophilic substitution mechanism for the hydrolysis of a
generic primary organic phosphate (RCH2OP + H2O → RCH2OH + Pi ), and show how
researchers, by running the reaction in H218O, (isotopically labeled water), could
potentially distinguish between a nucleophilic substitution and phosphate group transfer
mechanism by looking at where the 18O atom ends up in the products.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
27
Chapter 9: Phosphate transfer
Section 9.7: Phosphate diesters in DNA and RNA
Phosphate diesters play an absolutely critical role in nature - they are the molecular 'tape'
that connect the individual nucleotides in DNA and RNA via a sugar-phosphate
backbone. Take note of the 1' - 5' carbon numbering shown below for the ribose sugar these numbers will be used frequently in the coming discussion. The 'prime' symbol (') is
used to distinguish the ribose carbon numbers from the nucleotide base carbon numbers
(which are not shown here).
DNA
O
4'
O
R1
3'
1'
2'
O
=
O P O
O
Base
O
5'
R2
O P O
O
O
Base
phosphate diester
HO
fig 5
The introduction to this chapter referenced a widely-read 1987 commentary in Science
Magazine, in which F.H. Westheimer of Harvard University addressed the question of
why phosphates were 'chosen' by nature for critical biochemical job of linking DNA
(Science 1987, 235, 1173). He emphasizes how critical it is for the phosphate diester
linkage in DNA to be stable in water – in other words, it must be resistant to spontaneous
(nonenzymatic) hydrolysis. Even very infrequent occurrence of such an undesired
hydrolysis event could be disastrous for an organism, given that an intact DNA strand is a
long-term storage mechanism for genetic information.
Exercise 9.12: Draw a mechanism for the spontaneous hydrolysis of a phosphate diester.
Westheimer pointed out that the inherent stability of DNA is a due in large part to the
negative charge on the non-bridging oxygen of the phosphate diester linker, which
effectively repels nucleophilic water molecules and shields the electrophilic phosphorus
atom from attack.
28
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
H
H
O
DNA
X
O
O P O
negative charge protects
phosphorus from attack by
nucleophilic H2O electrons
O
DNA
fig 40
While DNA is quite stable with regard to spontaneous hydrolysis, it of course can be
degraded by specific enzymatic hydrolysis, where the phosphate electrophile is activated
for attack through noncovalent interactions (eg. with Mg2+) in the active site. Enzymes
that hydrolyze the phosphate diester bonds in DNA are called nucleases, and we will
learn more about them in section 9.8.
Unlike DNA, RNA is quite vulnerable to spontaneous hydrolysis in aqueous solution.
This does not present a physiological dilemma, because the function of RNA is to encode
genetic information on a temporary rather than long-term basis. Why does hydrolysis
occur so much more rapidly in RNA than in DNA? The answer has everything to do
with the lowered entropic barrier to the reaction (you might want to quickly review the
concept of entropy at this point). RNA nucleotides, unlike the deoxynucleotides of DNA,
have a hydroxyl group at the neighboring 2' carbon. The 2' hydroxyl group is right next to
the electrophilic phosphorus atom, poised in a good position to make a nucleophilic
attack, breaking the RNA chain and forming a cyclic phosphate diester intermediate (see
figure below).
RNA
RNA
RNA
O
O
O
O
Base
3' 2'
O
O
O
O
O
P O
O
O
O
O
:B
A
H
RNA
Base
3' 2'
3' 2'
O P O H
RNA
O
Base
O
O
cyclic phosphate diester
P
O
OH
RNA
fig 41
Researchers working with RNA have to be careful to store their samples at very cold
temperatures, preferably freeze-dried or precipitated in ethanol, to avoid hydrolysis. The
problem of RNA decomposition is compounded by the fact that RNAase enzymes, which
catalyze RNA hydrolysis, are present on the surface of human skin and are very stable,
long-lived, and difficult to destroy.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
29
Chapter 9: Phosphate transfer
In contrast, DNA samples can be safely stored in aqueous buffer in a refrigerator, or in a
freezer for longer-term storage.
Section 9. 8: The organic chemistry of genetic engineering
Many enzymes that catalyze reactions involving the phosphate diester bonds of DNA
have been harnessed for use in genetic engineering - techniques in which we copy, snip,
and splice DNA in order to create custom versions of genes. The tools of genetic
engineering have become indispensable and commonplace in the past decade, and most
researchers working on the biological side of chemistry use them extensively. The days
of painstakingly purifying an enzyme from bacterial cultures or ground-up cow livers are
pretty much gone. Now scientists clone the gene that encodes the enzyme, make any
desired changes (by site-directed mutagenesis, for example), and use a host such as E.
coli or yeast to produce 'recombinant' enzyme from the cloned gene. You will learn the
details of many of these procedures in a biochemistry or molecular biology course. What
we will focus on now is applying what we have learned about phosphate group transfer
reactions so that we can recognize some of the organic chemistry that is happening in a
cloning experiment.
The first thing you have to do in a gene cloning procedure is to copy a DNA strand. This
is accomplished by an enzyme called DNA polymerase (EC 2.7.7.7), which uses a single
strand of DNA as a template to synthesize a second, complementary strand (the full
picture of this complex process is well beyond the scope of this book, but recall that we
talked about the discovery of thermostable DNA polymerase in the introduction to
chapter 6).
You may have learned in a biology class that DNA is synthesized in the 3' to 5' direction.
Notice below that the 3' hydroxyl group on the end of the growing DNA strand attacks
the α-phosphate of a 2'-deoxynucleoside triphosphate (dNTP), expelling inorganic
pyrophosphate.
DNA polymerase reaction:
DNA
DNA
O
O
3' end of growing
DNA polymer
O
5'
Base
3'
3'
+
O
O
PPi
O
H
O
O
O P O
O
O P O P O P O
O
O
O
5'
O
3'
dNTP
Base
DNA polymer has extended
by one nucleotide
HO
30
Base
O
5'
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
5'
O
3'
HO
Base
Chapter 9: Phosphate transfer
fig 42
Scientists are able to cut DNA using 'molecular scissor' enzymes called restriction
endonucleases that cleave double-stranded DNA by hydrolysis at specific base
sequences.
DNA hydrolysis by restriction endonucleases:
DNA
DNA
O
O
Base
O
5'
3'
3'
HO
O
O P O
O
H
O
Base
O
5'
5'
O
O
Base
O P O
3'
H
5'
O
O
O
Base
3'
O
DNA
DNA
fig 43
Notice that the result of this hydrolytic cleavage reaction is one segment of DNA with a
hydroxy group at the 3' position, and a second segment with a phosphate group at the 5'
position.
A commonly used restriction endonuclease called 'BamHI' cleaves double-stranded DNA
specifically at the following 6-base sequence:
cut
5'
3'
G-G-A-T-C-C
C-C-T-A-G-G
3'
5'
cut
HO
5'
3'
OP
G G-A-T-C-C
C-C-T-A-G G
PO
3'
5'
OH
fig 43
Notice that a 'staggered' cut is made: this is a common (and useful) property of many
endonucleases, although some make 'blunt-ended' cuts.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
31
Chapter 9: Phosphate transfer
While an endonuclease cleaves a phosphodiester linkage in a DNA strand, DNA ligase
(EC 6.5.1.1) accomplishes the reverse process: it catalyzes the formation of a new 3'-5'
link between two strands:
DNA ligase reaction:
DNA1
O
O
5'
DNA1
Base
O
3'
O
5'
HO
Base
3'
O
O
O P O
O P O
O
5'
O
Base
AMP
PPi
ATP
new DNA linkage
5'
O
O
3'
Base
3'
O
O
DNA2
DNA2
fig 45a
Note that there is initially no leaving group on the 5' phosphate of DNA2, which makes a
direct phosphate transfer reaction impossible. The strategy employed by the DNA ligase
enzyme is to first activate the 5' phosphate of DNA2 using ATP (phase 1 below), then the
ligation reaction can proceed (phase 2)
DNA ligation steps
Phase 1: activation
O
O P O
O
OAMP
no leaving group: unactivated
5'
O
Base
ATP
PPi
activated for phosphoryl transfer
O P O
O
5'
3'
32
O
3'
O
O
DNA2
DNA2
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Base
Chapter 9: Phosphate transfer
Phase 2: ligation:
DNA1
O
O
5'
DNA1
Base
3'
O
HO
OAMP
activated for phosphoryl transfer
5'
O
Base
3'
AMP
O P O
O
O
5'
O
Base
O P O
5'
O
3'
O
Base
3'
O
O
new DNA linkage
DNA2
DNA2
One more enzymatic tool in the genetic engineering arsenal bears mention. In some
cloning procedures, a researcher may want to prevent unwanted ligation of DNA. This
can be accomplished by using the enzyme alkaline phosphatase (EC 3.1.3.1), which
catalyzes the dephosphorylation of many different organic phosphates, including 5'phosphorylated DNA (recall that we discussed phosphatases in section 9.6).
Alkaline phosphatase reaction:
H
O
H
O
no 5' phosphate ligation cannot occur
could be ligated to 3' end
of another DNA strand
HO
O P O
O
5'
O
3'
5'
Base
O
Base
3'
phosphatase
O
O
DNA
DNA
O
+
O P O
O
fig 46
With the phosphate group removed, ligation is impossible - there is no way to make a
new phosphodiester bond without a 5' phosphate group!
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
33
Chapter 9: Phosphate transfer
Section 9.9: NMR of phosphorylated compounds
Because so many biological molecules contain phosphoryl groups, it is worthwhile to
look at how scientists use NMR to determine the structure of these molecules. Recall
from section 5.1 that 31P, the most abundant isotope of phosphorus, is NMR active: it can
be directly observed by 31P-NMR, and indirectly observed in 1H-NMR and 13C-NMR
through its spin-coupling interactions with neighboring protons and carbons, respectively.
Consider the case of isopentenyl diphosphate, the building block molecule used by cells
to make 'isoprenoid' compounds such as cholesterol (in many animals), or β-carotene (in
some plants). NMR spectra of this molecule were taken in a D2O solvent, buffered with
ND4OD (the deuterium equivalent of aqueous ammonium hydroxide, NH4OH) (J. Org.
Chem. 1986, 51, 4768). In our discussion, carbon atoms are specified with numbers,
protons with lower case letters, and phosphorus atoms with upper case letters.
Hd
Ha 5 CH3
C 2 C
O PB O PA O 1 C 3 CH2
4
Hb Hb
O
O
O
O
Ha
Hc
First, let's look at the proton spectrum:
1
H-NMR
Ha: 4.05 ppm (td); 3JHa-Hb = 6.6 Hz; 3JHa-PA = 3.3 Hz.
Hb: 2.39 ppm (t) 3JHa-Hb = 6.6 Hz
Hc: 4.86 ppm (s)
Hd: 1.77 ppm (s)
Ha split into doublet
O
O
Ha
Ha
Ha split into triplet
O PB O PA O
O
O
Hb
Hb
The signals for Hb, Hc, and Hd look like we would expect from our discussion in chapter
5, with the exception of Hc which you will be invited to discuss in the exercise below.
Why, though, is the signal for Ha split into a triplet of doublets (td)? First of all, as,
expected, the two neighboring Hb protons split the Ha signal into a triplet, with 3JH-H = 6.6
Hz. Then, the signal is further split into doublets (3JH-P = 3.3 Hz) by PA, the closer of
34
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
the two phosphorus atoms. A phosphorus atom will spin-couple with protons up to three
bonds away.
Exercise 9.13: The signal for the two 'Hc' protons in isopentenyl diphosphate is reported
above as a singlet integrating to 2H. Are these two protons really chemically equivalent,
and, according to what you know about proton NMR, should this signal really be a
singlet? Explain what is going on.
Now, let's look at the 13C spectrum of IPP:
13
C-NMR (proton-decoupled)
C1: 40.7 ppm (d); 2JC1-PA = 7.2 Hz
C2: 67.0 ppm (d); 3JC2-PA = 4.0 Hz
C3: 147.4 ppm
C4: 114.6 ppm
C5: 24.5 ppm
C1 split into a doublet 5
O
O
O PB O PA O
O
O
1
2
3
4
C2 split into a doublet
Notice that the signals for both C1 and C2 are split into doublets by the magnetic field
of PA. Phosphorus atoms will spin-couple with 13C nuclei up to three bonds away.
Notice also that the 2-bond coupling between C1 and PA is larger than the 3-bond
coupling between C2 and PA (7.2 Hz vs. 4.0 Hz). Finally, notice that we do not observe 4bond C-P coupling: C3 is not spin-coupled to PA, and PB is not coupled to any of the 13C
or 1H nuclei on the molecule.
Remember that when processing a typical 13C-NMR spectrum, we electronically 'turn off'
spin coupling between carbons and neighboring protons in order to simplify the spectrum
(this is referred to as 'proton decoupling'). Proton decoupling does not turn off C-P spin
coupling.
Because 31P is NMR-active, we can also, with an NMR spectrophotometer equipped with
a phosphorus probe, directly observe the phosphorus NMR signals, just as we can
directly observe the signals from protons and 13C nuclei. On an NMR instrument where
protons resonate at 300 MHz and 13C nuclei resonate at 75 MHz, phosphorus resonates at
32 MHz. In 31P-NMR experiments, the reference standard used to determine the 0 ppm
point is usually phosphoric acid (tetramethylsilane, the standard 0 ppm point for 1H- and
13
C-NMR, doesn't have a phosphorus atom!). The 31P-NMR spectrum of isopentenyl
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
35
Chapter 9: Phosphate transfer
diphosphate has, as expected, two peaks, each of which is upfield of the phosphoric acid
standard (negative chemical shifts!) and split into a doublet (2JP-P = 20 Hz) due to 2-bond
coupling between the two phosphorus nuclei.
O
O
O PB O PA O
O
O
fig 13
PA: (-)11.03 ppm (d, 2JP-P = 20 Hz)
PB: (-)7.23 ppm (d, 2JP-P = 20 Hz)
Notice that although the C1 and C2 signals were split by PA in our 13C-NMR spectrum, in
the 31P-NMR spectrum the converse is not true: the PA signal is not split by C1 or C2.
Both of these carbons are NMR-inactive 12C isotope in 99 out of 100 molecules. In
addition, P-H splitting is not observed in this 31P spectrum, because proton decoupling is
in effect.
36
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
Key concepts to review
All of the reactions detailed in this chapter involved the transfer of a phosphate group usually a phosphate, diphosphate, or AMP group - from one molecule (the donor) to
another (the acceptor). Your learning goal for this chapter is to recognize and understand
what is happening in these phosphate group transfer reactions, and to gain a basic
understanding of the chemistry of phosphate and other phosphate groups.
Be sure that you can identify and provide examples of the following terms:
inorganic phosphate
inorganic phosphate
inorganic pyrophosphate
organic (mono)phosphate
organic diphosphate
organic triphosphate
phosphate (mono)ester
phosphate diester
phosphate anhydride
bridging/non-bridging oxygen
Also, make sure that you can recognize and use appropriately the various abbreviations
introduced in this chapter:
Pi
R-OPP
ADP
PPi
R-OAMP
AMP
R-OP
ATP
. . . in addition to the various structural abbreviations for adenosine mono-, di-, and
triphosphate.
You should know the approximate pKa values for phosphoric acid and an organic
monophosphate, and be able to state the approximate net charge (to the nearest 0.5 charge
unit) of these species in buffers of different pH levels.
You should be able to describe, in words and pictures, the tetrahedral sp3d bonding
picture for the phosphorus atom of a phosphate group. Even though the geometry is not
always shown in every drawing, always keep in mind that the phosphate group is
tetrahedral.
You should be able to draw resonance contributors for different phosphate groups,
identify major versus minor contributors, and explain why some are major and some are
minor. Remember - charges are shared between non-bridging oxygens, even if they are
not drawn that way!
Absolutely critical to your success with this chapter is being able to picture and
illustrate the mechanistic pattern which we refer to as a phosphate group transfer.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
37
Chapter 9: Phosphate transfer
Though not usually included in reaction illustrations, always remember that chargecharge interactions with magnesium ions and hydrogen bonds to active site amino acids
both serve to increase the electrophilicity of a phosphorus atom in donor compounds such
as ATP.
You should be able to identify the apical and equatorial positions in the pentavalent
intermediate of a phosphate transfer reaction, and recognize that the reaction results in
inversion at the phorphorus center.
You should be able to identify the α, β, and γ phosphate groups of ATP, as well as the
ribose and adenosine parts of the molecule.
You should be able to explain how ATP acts as a phosphate group donor, and why such
reactions are thermodynamically favorable.
You should be familiar with a variety of reactions in which ATP acts as a phosphate
group donor: given the starting compounds and products, you should be able to propose a
reasonable curved-arrow mechanism, including the structure of the pentavalent
intermediate. You should be able to predict the result of nucleophilic attack at the α, β, or
γ phosphates of ATP.
In general, you should be able to propose a likely mechanism for any phosphate transfer
reaction, given the starting compounds and products.
Given information about the existence of a covalently linked enzyme-substrate complex
in an enzyme mechanism, you should be able to propose a likely mechanism that
accounts for such an intermediate. For example, after being told that the phosphotyrosine
phosphatase reaction involves a phosphocysteine intermediate, you should be able to
propose a mechanism such as that shown in figure 38c.
38
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
Problems
In all of the problems that follow, feel free to use appropriate abbreviations when
drawing structures. However, always be sure not to abbreviate regions of a structure
which are directly involved in bond-breaking or bond-forming events.
P9.1: Draw a likely mechanism for reaction catalyzed by shikimate kinase (EC 2.7.1.71)
in the aromatic amino acid biosynthesis pathway). Stereochemistry of the product is not
indicated in the figure below - in your mechanism, show the stereochemistry of the
product, and explain how you are able to predict it from your knowledge of kinase
reactions.
CO2
CO2
ATP
HO
ADP
(stereochemistry not
indicated)
PO
OH
OH
OH
OH
P9.2: Draw a likely mechanism for the following reaction (EC 2.7.2.3) in the
gluconeogenesis pathway, and predict what compound is indicated by the question mark.
ATP
OH
O
OP
O
?
OH
PO
OP
O
P9.3:
a) Draw a likely mechanism for the following reaction (EC 4.3.6.2) from ribonucleotide
biosynthesis. Hint: what is the nucleophilic group? How could the enzyme increase it's
nucleophilicity?
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
39
Chapter 9: Phosphate transfer
O
H
N
O
PPO
OP
ATP
N
N
ADP
O
O
PPO
N
O
OH OH
OH OH
b) Draw a mechanism for the following reaction, also from ribonucleotide biosynthesis
(EC 6.3.3.1):
OP
O
OP
NH
H
N
O
OP
ATP
ADP
O
NH
NH
HO
N
H
N
HO
OH
OH
P9.4:
a) The carboxylate group on the amino acid valine is activated in an early step in the
biosynthesis of the antibiotic penicillin. Predict the product of this reaction, and draw the
likely intermediate of the phosphate group transfer reaction.
ATP
H3N
PPi
CO2
valine
P9.5: The reaction below is an early step in the synthesis of tyvelose, a sugar found on
the surface of some pathogenic bacteria. Notice that CTP plays the role of the phosphate
group donor in this case, rather than ATP.
OH
HO
HO
40
CTP
O
HO
OH
?
HO
HO
OP
O
HO
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OCDP
Chapter 9: Phosphate transfer
Draw a mechanism for the reaction, and indicate the second product that is released by
the enzyme. (J. Biol. Chem. 2005, 280, 10774)
P9.6: Draw the likely mechanism and product of the following hypothetical phosphate
group transfer reactions. Hint - think about which phosphate group of ATP is the
electrophile in each case.
a)
ATP
Pi
ATP
AMP
R OH
b)
R OH
P9.7: The figure below illustrates an experiment in which a reaction catalyzed by an E.
coli enzyme was run in isotopically labeled water.
OH
18OH
HO
HO
2
OH
GDP
O
HO
HO
HO
OGDP
O
18
OH
HO
a) The researchers concluded that the reaction was not a phosphate group transfer.
Explain their reasoning.
b) Draw the products that would be expected if the reaction actually did proceed by a
phosphate transfer mechanism (be sure to show stereochemistry and the location of the
18
O atom). (Biochemistry 2000, 39, 8603)
P9.8: The reaction below proceeds with a direct attack by a water molecule on the
substrate, but the hydrolysis could be expected to proceed through two possible
mechanisms. Draw two possible mechanisms for the reaction run in H218O. Trace the
progress of the 18O 'label' throughout each mechanism to see where it ends up: this should
indicate to you how the two mechanisms could be (and in fact were!) distinguished
experimentally.
O
ADP
O
HO
OP
OH
OH
H2O
AMP
O
HO
OH
OH
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
41
Chapter 9: Phosphate transfer
P9.9: Glucose-6-phosphate is dephosphorylated to glucose in the last step of the
gluconeogenesis pathway. The reaction is not a direct hydrolysis: like the
phosphotyrosine phosphatase reaction we saw in this chapter it involves formation of a
phosphoenzyme intermediate, but in this case the enzyme residue acting as the initial
phosphate acceptor is an active site histidine rather than an asparate. Given this
information, propose a likely mechanism for the reaction.
P9.10: (This question assumes a basic knowledge of DNA structure and the idea of
supercoiling). DNA topoisomerase enzymes catalyze the temporary 'nicking' of one
strand of double-stranded DNA, which allows supercoiled DNA to 'unwind' before the
nicked strand is re-ligated. During the unwinding process, the 5' end of the nicked strand
is transferred to a tyrosine in the enzyme's active site, effectively holding it in place while
the 3' end rotates. Overall, the stereochemical configuration of the bridging phosphate is
retained. Propose a likely mechanism for this nicking and re-ligating process.
(Biochemistry 2005, 44, 11476.)
P9.11: Pictured below is a series of phosphate group transfer steps in the early part of
isoprenoid biosynthesis in bacteria. With the knowledge that the atoms in green are
derived from ATP, predict the structures of compounds A, B and C. (EC 2.7.7.60, EC
2.7.1.148, EC 4.6.1.12)
from ATP
CTP
ATP
PPi
ADP
B
A
O O
O P
O
CMP
C
O
OH
O
P
O
OH
P9.12: The reaction below shows the synthesis of glucose-UDP, an important
intermediate in carbohydrate biosynthesis. Notice that UTP (instead of ATP) is the
phosphate donor. Identify the by-product denoted below by a question mark. (Molecules
and Cells 2010, 29, 397; E.C. 2.7.7.9)
OH
HO
HO
UTP
O
OH
?
HO
HO
OH
OP
O
OH
OUDP
P9.13: Isomerization of 3-phosphoglycerate to 2-phosphoglycerate (a reaction in
glycolysis) has been shown to occur with the participation of a phosphohistidine residue
in the enzyme's active site. The two phosphate groups are distinguished in the figure
42
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 9: Phosphate transfer
below by color. With this information, propose a mechanism for the reaction. (From J.
Mol. Biol. 1999, 286, 1507)
enz
enz
O
O
O
P
NH
N
P
O
O
O
N
NH
O
O
P
O
O
OH
O P O
O
O
O
HO
O
O
O
2-phosphoglycerate
3-phosphoglycerate
P9.14: The gluconeogenesis (sugar-building) pathway enzyme glucose-6-phosphatase
catalyzes an indirect phosphate hydrolysis reaction with a phosphohistidine intermediate
('indirect hydrolysis' in this context means that a water molecule does not directly attack
glucose-6-phosphate).
OP
HO
HO
H2O
O
OH
OH
glucose-6-phosphate
OH
Pi
O
HO
HO
OH
OH
glucose
Researchers wanted to confirm that the hydrolysis in this reaction is indirect, rather than
direct. It turns out that the same enzyme is also capable of catalyzing the transfer of the
phosphate group from glucose-6-phosphate to the hydroxyl group on carbon #6 of
another glucose molecule (instead of to water, which is the natural reaction). The
enzyme-catalyzed transfer of phosphate between two glucose substrates is reversible.
The researchers incubated the enzyme with labeled glucose-6-phosphate, in which in the
phosphate center was chiral (with the R configuration) due to the incorporation of 17O
and 18O isotopes. They also included a high concentration of glucose in the reaction
mixture, which ensured that the glucose-to-glucose transfer reaction predominated and
hydrolysis (the 'natural' reaction) did not take place. After allowing the reaction to reach
equilibrium, they isolated the glucose-6-phosphate and looked at the configuration of the
phosphate group.
Given what you have just learned about the enzyme mechanism, predict what the
researchers found in this experiment, explain your prediction, and draw the appropriate
structure(s), including stereochemistry. Assume that the glucose-to-glucose mechanism is
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
43
Chapter 9: Phosphate transfer
identical to the hydrolysis mechanism, aside from the identity of the ultimate phosphate
acceptor. Biochem J., 1982, 201, 665.
44
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Chapter 10
Nucleophilic addition to carbonyl groups
Introduction
It's possible that the fuel for the car you drive thirty years from now will come from the
back end of a panda. Not literally, of course – but it just might turn out that future biofuel
technology will be derived in part from the stuff that workers have to clean out of the
enclosure housing Ya Ya and Le Le, the two resident pandas at the Memphis Zoo in . At
least, that's the hope of Dr. Ashli Brown, a biochemistry professor at Tennessee State
University.
First, a little background. If you are like most people in the United States, you are already
burning ethanol every time you drive: in 2012, the U.S. Department of Energy reports
that over 13 million gallons of ethanol were sold at gas stations nationwide, most often as
a 10% mixture along with 90% conventional gasoline. The ethanol we burn today is
made by fermenting the sugars present in edible corn. The use of corn ethanol, while a
significant step forward in the effort to move away from petroleum fuels and towards
carbon-neutral, renewable energy sources, is far from a permanent, sustainable solution to
the world's ever-increasing energy needs. Growing corn crops requires a lot of energy
and expense, from running the large equipment used to plow and harvest the fields, to
manufacturing and applying pesticides and fertilizers, all the way to trucking the corn to
the ethanol plant. In fact, some calculation methods suggest that more energy goes into
producing a gallon of corn-based ethanol than is released when the ethanol is burned.
Moreover, growing corn requires a lot of water, and takes up land which otherwise could
be used for growing food, or preserved as a natural habitat. A recent study by scientists in
South Dakota reported that between 2006 and 2011, a full 1.3 million acres of wetland
and prairie were plowed over and converted to biofuel crop production in five
midwestern states.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
45
Chapter 10: Nucleophilic addition to carbonyls
What would be much better in the long run is if we could produce ethanol or other
biofuels not from resource-intensive food crops like corn, but from non-edible plant
materials: grasses, trees, and agricultural byproducts such as the cobs and stalks from
corn plants. Switchgrass, for example, is a native North American prairie grass that is
thought to have high potential for biofuel production.
So if we can make ethanol from corn, couldn't we just change over to switchgrass using
the same technology?
Unfortunately, it's not nearly that simple. Ethanol is made by 'feeding' glucose to living
yeast cells, allowing them to break down the sugar into ethanol – a metabolic process
called fermentation. Corn kernels contain sugar in the form of starch, a polysaccharide of
linked glucose molecules. Enzymes called 'amylases' are used to break up the starch
polymer into individual glucose molecules (as well as two-glucose units called
cellobiose), which are then fermented by the yeast.
OH
note stereochemistry here
O
O
H
HO
OH
O
OH
HO
O
section of starch polymer
HO
O
fig1b
The rest of the corn plant – the stalks, leaves, and cobs – is composed in large part of
another glucose polymer called cellulose.
OH
note stereochemistry here
O
O
HO
OH
HO
O
O
OH
H
H
O
OH
section of cellulose polymer
fig 1c
Cellulose is a major component of plant cell walls, and is the most abundant organic
compound on the planet - an enormous source of glucose for fermentation! The problem,
from a renewable energy perspective, is how to get at the glucose monomers that make
46
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
up the polymer. Look closely at the bond connecting two glucose monomers in starch,
and then compare it to the same bond in cellulose. They both link the same two carbons
of glucose, but with opposite stereochemistry. Recall that enzymes are very sensitive to
the stereochemical configuration of their substrate molecules. It should come as no
surprise, then, that the amylase enzymes which are so efficient at breaking apart starch
are completely ineffective at breaking apart cellulose. Other enzymes, known as
cellulases, are needed for this job. These enzymes do exist in nature: just think about
what happens to tree branches, leaves, and other cellulose-rich plant matter that lies on
the forest floor. These slowly rot away, the cellulose broken apart by cellulase enzymes
in microscopic fungi.
The key word here, though, is 'slowly'. Fungi living on the forest floor are not in any
great hurry to degrade the leaves and wood around them – the cellulose is not going
anywhere. Fungal cellulases are, comparatively speaking, very slow, inefficient enzymes.
Herein lies the biggest challenge to the development of economically viable production
of ethanol from cellulosic sources such as switchgrass or wood. Breaking the glucoseglucose bonds in cellulose is the main bottleneck in the whole process.
This is where the pandas come in.
Pandas live primarily on a diet of bamboo, obtaining their energy from the cellulose in
the plant. Like other plant-eaters such as cows, horses, and sheep, pandas do not make
their own cellulase enzymes. Rather, they rely on a diverse population of symbiotic
microbes inhabiting their digestive tracts to do the job of cellulose digestion for them.
Unlike the microbes living the slow-paced lifestyle of the forest floor, though, the panda's
microbes don't have a lot of time to spare - the food is moving through the system pretty
quickly. In theory, evolutionary pressure should have resulted in panda-gut microbes
with speedy cellulase enzymes, and that is what Dr. Ashli Brown at Tennessee State was
hoping to find as she and her research students analyzed panda feces from the Memphis
Zoo. They have had some success: at the fall, 2013 meeting of the American Chemical
Society, Dr. Brown announced that her group, working in cooperation with colleagues at
the University of Wisconsin, had found over forty cellulose-digesting bacteria, courtesy
of Ya Ya and Le Le. The next step is to clone the cellulase- encoding genes, use the
DNA to produce recombinant enzyme, and see just how fast they are.
Other less cuddly and photogenic animals are also being studied with similar goals in
mind. Dr. Falk Warnecke, working at the U.S. Department of Energy Joint Genome
Institute in Northern California, has been investigating the microbes that live in the guts
of wood-eating termites, and many other researchers around the world are interested in
the symbiotic bugs which inhabit the rumen of cows and sheep.
The problematic chemical reaction catalyzed by cellulase enzymes is, in organic
chemistry terminology, an 'acetal hydrolysis'. Acetals are derived from aldehydes. The
reactions that occur at the carbonyl carbon of aldehydes and ketones is absolutely central
to the chemistry of carbohydrates such as starch and cellulose, and it is this chemistry that
is the subject of the chapter we are about to begin.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
47
Chapter 10: Nucleophilic addition to carbonyls
Section 10.1 Nucleophilic addition to aldehydes and ketones: an overview
10.1A: The aldehyde and ketone functional groups
Recall from chapter 1 that the ketone functional group is made up of a carbonyl bonded
to two carbons, while in an aldehyde one (or both) of the neighboring atoms is a
hydrogen.
O
O
R
C
R
R
ketone
C
H
aldehyde
fig1d
You probably are familiar with the examples shown below: acetone, the simplest ketone
compound, is the solvent in nail polish remover, benzaldehyde is the flavoring in
maraschino cherries, and formaldehyde (a special case in which the carbonyl carbon is
bonded to hydrogens on both sides) is the nasty-smelling stuff that was used to preserve
the unlucky frog that you dissected in high school biology class. The male sex hormone
testosterone contains a ketone group in addition to alcohol and alkene groups.
CH3
O
O
H3C
CH3
O
H
CH3
acetone
H
benzaldehyde
OH
H
O
formaldehyde
testosterone
fig 1e
Recall from chapter 2 the bonding picture in a ketone or aldehyde: the carbonyl carbon is
sp2 hybridized, with its three trigonal planar sp2 orbitals forming σ bonds with orbitals on
the oxygen and on the two carbon or hydrogen atoms. The remaining unhybridized 2p
orbital is perpendicular to the plane formed by the sp2 orbitals, and forms a π bond
through a side-by-side overlap with a 2p orbital on the oxygen. The σ and p bonds
between the carbon and oxygen combine to make the C=O double bond that defines the
carbonyl functionality.
48
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
R
R
C
=
C O
R
blue: sp2 orbital σ overlap
O
red: 2p orbital π overlap
R
fig 1
10.1B: Nucleophilic addition
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so
electron density is higher on the oxygen end of the bond and lower on the carbon end.
Recall that bond polarity can be depicted with a dipole arrow (A in the figure below), or
by showing the oxygen as bearing a partial negative charge and the carbonyl carbon a
partial positive charge (B).
O
R
C
R
O δ−
C δ+
R
R
O
R
C
O
R
R
major
A
B
C
R
minor
C
fig 2
A third way to illustrate the carbon-oxygen dipole (C in the figure above) is to consider
the two main resonance contributors: the major form, which is what you typically see
drawn in Lewis structures, and a minor but very important contributor in which both
electrons in the π bond are localized on the oxygen, giving it a full negative charge. The
latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
However the bond polarity is depicted, the end result is that the carbonyl carbon is
electron-poor - in other words, it is an electrophile. In addition, the trigonal planar
geometry means that the carbonyl group is unhindered). Thus, it is an excellent target for
attack by an electron-rich nucleophilic group, a mechanistic step called nucleophilic
addition:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
49
Chapter 10: Nucleophilic addition to carbonyls
Nucleophilic addition to an aldehyde or ketone (enzymatic)
active site acid poised to protonate oxygen
A
H
A
enz
R
Nu
H
O
O
C
C
R
R
Nu
H
:B
OH
Nu
R
C R
R
H
enz
:B
active site base poised to deprotonate nucleophile
as it bonds to the electrophilic carbon
fig 3
Notice the acid-base catalysis that is going on in this generalize mechanism: in the
enzyme active site, a basic group is poised to deprotonate the nucleophile (thus enhancing
its nucleophilicity) as begins to attack the carbonyl carbon, while at the same time an
acidic proton on another active site group is poised just above the carbonyl oxygen (thus
enhancing the electrophilicity of the carbon), ready to protonate the oxygen and
neutralize any negative charge that builds up.
10.1C: Stereochemistry of nucleophilic addition to a carbonyl
Recall from section 3.11B that when the two groups adjacent to a carbonyl are not the
same, we can distinguish between the re and si 'faces' of the planar structure.
The concept of a trigonal planar group having two distinct faces comes into play when
we consider the stereochemical outcome of a nucleophilic addition reaction. Notice that
in the course of a carbonyl addition reaction, the hybridization of the carbonyl carbon
changes from sp2 to sp3, meaning that the bond geometry changes from trigonal planar to
tetrahedral. If the two R groups are not equivalent, then a chiral center is created upon
addition of the nucleophile. The configuration of the new chiral center depends upon
which side of the carbonyl plane the nucleophile attacks from.
50
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
attack at re face:
attack at si face:
H A
Nu
H
H A
O
OH
OH
O
C
R2 R1
C R2
Nu
R1
R2 C
Nu
R1
C
B:
R2 R1
H Nu
:B
enantiomers
(assume R1 is higher priority than R2)
fig 4
If the reaction is catalyzed by an enzyme, the stereochemistry of addition is (as you
would expect) tightly controlled, and leads to one stereoisomer exclusively- the
nucleophilic and electrophilic substrates are bound in specific positions within the active
site, so that attack must occur specifically from one side and not the other. Nonenzymatic
reactions of this type often result in a 50:50 mixture of stereoisomers, but it is also
possible that one stereoisomer may be more abundant, depending on the structure of the
reactants and the conditions under which the reaction takes place. We'll see some
examples of this phenomenon soon when we look at cyclic forms of sugar molecules.
Section 10.2: Hemiacetals, hemiketals, and hydrates
10.2A: Overview
One of the most important examples of a nucleophilic addition reaction in biochemistry,
and in carbohydrate chemistry in particular, is the addition of an alcohol to a ketone or
aldehyde. When an alcohol adds to an aldehyde, the result is called a hemiacetal; when
an alcohol adds to a ketone the resulting product is a hemiketal.
OH
O
R2OH
+
alcohol
R2OH
alcohol
R1
C
H
R1
C
H
OR2
aldehyde
hemiacetal
O
OH
+
R1
C
R3
ketone
R1
C R3
OR2
hemiketal
fig 5
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
51
Chapter 10: Nucleophilic addition to carbonyls
(The prefix ‘hemi’ (half) is used in each term because, as we shall soon see, addition of a
second alcohol nucleophile can occur, resulting in species called acetals and ketals.)
The conversion of an alcohol and aldehyde (or ketone) to a hemiacetal (or hemiketal) is a
reversible process. The generalized mechanism for the process at physiological pH is
shown below.
Biochemical mechanism of hemiacetal formation:
:B
H
H A
O
R1
C
O
:B
H
H OR2
alcohol
aldehyde
R1 C
H
OR2
H A
hemiacetal
fig 6
In general, hemiacetals (and hemiketals) are higher in energy than their aldehyde-alcohol
components, so the equilibrium for the reaction lies to the left. As we will soon see in the
context of glucose and other sugars, however, five- and six-membered cyclic hemiacetals
are considerably lower in energy, and are favored at equilibrium: recall from chapter 3
the inherent stability of five- and six-membered rings.
Aldehydes and ketones, when in aqueous solution, exist in equilibrium with their hydrate
form. A hydrate forms as the result of a water molecule adding to the carbonyl carbon of
the aldehyde or ketone.
O
R
C
OH
H
aldehyde
+
H2O
R1 C
H
OH
water
hydrate
fig 6a
Although you should be aware that aldehyde and ketone groups may exist to a
considerable extent in their hydrated forms when in aqueous solution (depending upon
their structure), they are usually drawn in their non-hydrated form for the sake of
simplicity.
The mechanism we just saw for hemiacetal formation applies to biochemical reactions
occurring at physiological pH. In the organic laboratory, however, hemiacetal and
hemiketal formation usually takes place in the presence of a strong acid. The acid
catalyzes the reaction by protonating the carbonyl oxygen, thus increasing the
52
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
electrophilicity of the carbonyl carbon. Notice in the mechanism below that highly acidic
intermediates are drawn which would be unreasonable to propose for the corresponding
biochemical mechanisms occurring at physiological pH.
Acid-catalyzed hemiacetal formation (non-biological)
these intermediates have pKa ~ 0, and are too acidic to
form in a biological environment
A
H
O
O
R1
C
H
R1
C
H
O
R1 C
H
A
HOR2
H
H
H
OR2
O
R1 C
H
H
OR2
fig 6b
10.2B: Sugars as intramolecular hemiacetals and hemiketals
As stated above, the reactions of hemiacetals and hemiketals are central to the chemistry
of carbohydrates. Recall that sugar molecules generally contain either an aldehyde or a
ketone functional group, in addition to multiple alcohol groups. Aldehyde sugars are
often referred to as aldoses; ketone sugars as ketoses. For example, glucose is an aldose,
and fructose is a ketose - their structures are drawn below in Fischer projection:
O
OH
OH
O
HO
HO
OH
OH
OH
OH
OH
OH
glucose
(an aldose)
fructose
(a ketose)
fig 8
Exercise 10.1: What term describes the relationship between glucose and fructose (in
other words, what kind of isomers are they)?
Glucose and fructose are shown above in their open-chain form. However, recall from
section 1.3C that in aqueous solution, glucose, fructose, and other sugars of five or six
carbons rapidly interconvert between straight-chain and cyclic forms. This occurs
through the formation of intramolecular hemiacetals and hemiketals. This simply means
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
53
Chapter 10: Nucleophilic addition to carbonyls
that the 'R' group of the alcohol is already covalently attached to the 'R 'group of the
aldehyde (R1 in our general mechanism).
alcohol group is on same
molecule as aldehyde
O
OH
OH
O
cyclic acetal
fig 9a
Unlike most of the biochemical reactions you will see in this text, sugar cyclization
reactions are not catalyzed by enzymes: they occur spontaneously and reversibly in
aqueous solution. For most five- and six-carbon sugars, the cyclic forms predominate in
equilibrium.
The cyclic form of glucose is a six-membered ring, with an intramolecular hemiacetal
formed by attack of the hydroxl on carbon #5 to the aldehyde carbon (carbon #1, also
called the anomeric carbon in carbohydrate terminology).
spontaneous in aqueous solution:
1
O
2
OH
HO
H water acts as catalytic base and acid
H
6
3
4
O
HO
OH
5
OH
6
OH
=
HO 4
HO
O
O
5
3
2
OH
H
1
open chain glucose
(<<1% at equilibrium)
HO
HO
HO
HO
OH
OH
O
HO
O
+
HO
OH
α-glucopyranose
β-glucopyranose
(37% in equilibrium)
(63% in equilibrium)
RO
R C OH
RO
R C H
H
OH
(cyclic hemiacetal)
fig 9
54
OH
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
The cyclic form of glucose is called glucopyranose. As was discussed above,
nucleophilic attack on a planar carbonyl group can occur at either side of the plane,
leading to two different stereochemical outcomes - in this case, to two different
diastereomers. In carbohydrate nomenclature, these two diastereomers are referred to as
the α and β anomers of glucopyranose.
Because the formation of glucopyranose occurs spontaneously without enzyme catalysis,
shouldn’t equal amounts of these two anomers form? In fact, this does not happen: there
is almost twice as much of one anomer than the other at equilibrium. Why is this?
Remember (section 3.2) that six-membered rings exist predominantly in the chair
conformation, and that the lower energy chair conformation is that in which unfavorable
interactions between substituents are minimized – in most cases, this is the conformation
in which larger substituents are in the equatorial position. In the lower-energy chair
conformation of the major β anomer of glucopyranose, all of the hydroxyl groups are in
the equatorial position, but in the minor α anomer one hydroxyl group is forced into the
axial position. As a result, the α anomer is higher in energy, and less abundant at
equilibrium.
HO
O
HO
HO
HO
HO
OH
OH
H
β-glucopyranose
(63% in equilibrium)
HO
HO
OH
O
OH
H
open-chain glucose
(<<1% at equilibrium)
O
HO
H
HO
OH
OH
α-glucopyranose
(37% in equilibrium)
fig 10
Exercise 10.2: Draw a mechanism for the conversion of α-glucopyranose to open-chain
glucose.
Fructose in aqueous solution forms a six-membered cyclic hemiketal called
fructopyranose when the hydroxyl oxygen on carbon #6 attacks the ketone carbon
(carbon #2, the anomeric carbon in fructose).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
55
Chapter 10: Nucleophilic addition to carbonyls
1
OH
2
O
HO
3
4
OH
5
OH
6
=
O
H
6
OH
O
OH
5
open-chain fructose (~1%)
HO
OH
4
2
1
OH
3
OH
O
OH
OH
HO
OH
α-fructopyranose (~1%)
OH
O
+
OH
OH
HO
OH
β-fructopyranose (70%)
fig 11
In this case, the β anomer is heavily favored in equilibrium by a ratio of 70:1, because in
the minor α anomer the bulkier CH2OH group occupies an axial position.
Notice in the above figure that the percentages of α and β anomers present at equilibrium
do not add up to 100%. Fructose also exists in solution as a five-membered cyclic
hemiketal, referred to in carbohydrate nomenclature as fructofuranose. In the formation
of fructofuranose from open-chain fructose, the hydroxyl group on the fifth carbon
attacks the ketone.
56
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
1
OH
2
O
HO
3
4
=
OH
5
OH
6
OH
HO
6
H
1
OH
O
5
4
HO
2
3
O
OH
open-chain fructose
HO
O
HO
OH
OH
O
OH
OH
+
HO
OH
β-fructofuranose (23%)
HO
OH
α-fructofuranose (5%)
fig 12)
In aqueous solution, then, fructose exists as an equilibrium mixture of 70% βfructopyranose, 23% β-fructofuranose, and smaller percentages of the open chain and
cyclic α-anomers. The β-pyranose form of fructose is one of the sweetest compounds
known, and is the main component of high-fructose corn syrup. The β-furanose form is
much less sweet.
Although we have been looking at specific examples for glucose and fructose, other fiveand six-carbon monosaccharides also exist in solution as equilibrium mixtures of open
chais and cyclic hemiacetals and hemiketals. Shorter monosaccharides are unlikely to
undergo analogous ring-forming reactions, however, due to the inherent instability of
three and four-membered rings.
Exercise 10.3:
a) Identify the anomeric carbon of each of the sugars shown below, and specify whether
the structure shown is a hemiacetal or hemiketal.
b) Draw mechanisms for cyclization of the open-chain forms to the cyclic forms shown
(for simplicity, start with the 'curled up' conformation of the open chain forms, as in figs
9-12).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
57
Chapter 10: Nucleophilic addition to carbonyls
HO
OH
O
HO
HO
HO
OH
OH
O
sorbose
HO
OH
ribose
fig 12a
Section 10.3: Acetals and ketals
10.3A: Overview
Hemiacetals and hemiketals can react with a second alcohol nucleophile to form an acetal
or ketal. The second alcohol may be the same as the first (ie. if R2 = R3 in the scheme
below), or different.
OH
R1 C
H
H
+
R3OH
R1 C OR2 + H2O
OR2
OR3
hemiacetal
acetal
fig 13
Although we focus here on biological reactions, it is instructive in this case to consider
non-biological acetal-forming reactions before we look at their biochemical counterparts.
In a non-enzymatic context, acetal/ketal formation - just like hemiacetal/hemiketal
formation - is generally catalyzed by a strong acid.
58
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Acid-catalyzed acetal formation (non-biological)
R1
C
H
OR2
A
H
resonance-stabilized
carbocation
protonation of OH makes it a
good leaving group
OH
R1 C
OH2
H
R1 C
OR2
R1
H
OR2
C
H
OR2
R3OH
hemiacetal
H
H
R1 C OR2
R1 C OR2
OR3
A
H
acetal
OR3
fig 18b
The role of the acid catalyst is to protonate the OH group of the acetal, thus making it a
good leaving group (water). Notice something important here: the conversion of a
hemiacetal to an acetal is simply an SN1 reaction, with an alcohol nucleophile and
water leaving group. The carbocation intermediate in this SN1 mechanism is stabilized by
resonance due to the oxygen atom already bound to the electrophilic carbon.
Below are some examples of simple, non-biological acetal and ketals.
O
OCH3
H3C C OCH3
O
O
O
O
O
H
A
B
C
D
fig
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
59
Chapter 10: Nucleophilic addition to carbonyls
13a
Exercise 10.4: For each acetal/ketal A-D in the figure above, specify the required
aldehyde/ketone and alcohol starting materials.
Exercise 10.5: Categorize each of the following molecules as a hemiacetal, hemiketal,
acetal, ketal, hydrate of an aldehyde, or hydrate of a ketone.
HO
OCH3
H
OH
O
O
O
A
B
C
OH
O
O
PO
OH
OH
D
E
fig 15c
Exercise 10.6: Specify the acetal/ketal that would form from a reaction between the given
starting compounds.
O
OH
a)
and
(2 molar equivalents)
b)
O
and
HOCH2CH2OH
(1 molar equivalent)
fig 15d
Exercise 10.7: Specify the aldehyde/ketone and alcohol combination that would be
required to form the compounds in figure 15c.
10.3B: Glycosidic bond formation
Now, let's consider acetal formation in a biochemical context. A very important example
of the acetal/ketal group in biochemistry is the glycosidic bonds which link individual
60
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
sugar monomers to form polysaccharides (see section 1.3C for a quick review). Look at
the glycosidic bond between two glucose monomers in a cellulase chain:
OH
acetal
6
cellulose
O 4
HO
O
HO
5
OH
1
5
4
6
H
R2O
R1 C
H
H
cellulose
1
2
O
2
3
OH
3
O
O
OH
OR3
fig 16
If you look carefully, you should recognize that carbon #1, the anomeric carbon on the
left-side glucose monomer, is the central carbon of an acetal group. Biochemists refer to
this as a β-1,4 linkage, because the stereochemistry at carbon #1 is β in the specialized
carbohydrate nomenclature system, and it is linked to carbon #4 of the next glucose on
the chain. The vast structural diversity of carbohydrates stems in large part from the
different linkages that are possible - both in terms of which two carbons are linked, and
also the stereochemistry of the linkage. You will see many more variations of glycosidic
bond linkage patterns if you study carbohydrate biochemistry in greater depth.
Reactions in which new glycosidic bonds are formed are catalyzed by enzymes called
glycosyltransferases, and in organic chemistry terms these reactions represent the
conversion of a hemiacetal to an acetal (remember that sugar monomers in their cyclic
form are hemiacetals and hemiketals). The mechanism for glycosidic bond formation in a
living cell parallels the acid-catalyzed (non-biological) acetal-forming mechanism, with
an important difference: rather than being protonated, the OH group of the hemiacetal is
converted to a good leaving group by phosphorylation (this is a pattern that we are
familiar with from chapters 9 and 10). The specific identity of the activating phosphate
group varies for different reactions, so it is generalized in the figure below.
Mechanism for (biochemical) acetal formation:
Hemiacetal activation phase:
OR2
OR2
H
R1 C
OH
hemiacetal
step A
(see text below)
R1 C
O
H
good leaving group
OR
P
O O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
61
Chapter 10: Nucleophilic addition to carbonyls
acetal formation phase:
O
OR2
R1 C
O
OR
O O
H
P
P
OR
O O
step 1
O
R2O
H
C
H
R1
R3
OR2
:B
step 2
R1 C OR3
H
acetal
resonance-stabilized
carbocation intermediate
R2O
C
H
R1
fig 18
Step A (Activation phase): This phase of the reaction varies according to the
particular case, but always involves phosphate group transfer steps that are
familiar from chapter 9. What is most important for our present discussion,
however, is simply that the hydroxyl group on the hemiacetal has been activated ie. made into a good leaving group - by phosphorylation.
Step 1: Now that the leaving group has been activated, it does its job and leaves,
resulting in a resonance stabilized carbocation.
Step 2: A nucleophilic alcohol on the growing cellulose chain attacks the highly
electrophilic carbocation to form an acetal. Here is where the stereochemistry of
the new glycosidic bond is determined: depending on the reaction, the alcohol
nucleophile could approach from either side of the planar carbocation.
To reiterate: it is important to recognize the familiar SN1 mechanistic pattern in play here:
in step A, a poor leaving group is converted into a good leaving group, in step 1 the
leaving group leaves and a stabilized carbocation is left behind, and in step 2 a
nucleophile attacks to form a new bond and complete the substitution process. Look
back at the SN1 reactions we saw in chapter 8 if you are having trouble making this
mechanistic connection.
Now, let's look specifically at the glycosyl transferase reaction mechanism in which a
new glycosidic bond is formed on a growing cellulose chain. Glucose (a hemiacetal) is
first activated through two enzymatic phosphate transfer steps: step A1, a phosphate
isomerization reaction with a mechanism similar to the reaction in problem P9.13,
followed by a UTP-dependent step A2, for which you were invited to propose a
mechanism in problem P9.12.
62
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
OP
OH
hemiacetal
O
HO
H
HO
OH
H
HO
step A1
OH
O
HO
poor leaving group
OH
OP
glucose-1-phosphate
glucose-6-phosphate
UTP
step A2
PPi
OH
activated hemiacetal
O
HO
H
HO
OH
O
UDP
good leaving group
glucose-UDP
fig 19a
The UDP group on glucose-UDP then leaves (step 1 below), forming a resonancestabilized carbocation intermediate. Attack by the alcohol group on the growing cellulose
chain in step 2 forms the glycosidic (acetal) bond. Note the inversion of stereochemistry.
Mol. Plant 2011, 4, 199
OH
activated hemiacetal
O
HO
HO
OH
UDP
H
OH
step 1
O
HO
HO
H
OH
UDP
cellulose
O
O
O
H
OH
resonance-stabilized
carbocation
glucose-UDP
OH
HO
O
growing end of cellulose polymer
:B
step 2
OH
acetal
O
HO
HO
OH
HO
H
cellulose
O
O
OH
H
O
OH
fig 19
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
63
Chapter 10: Nucleophilic addition to carbonyls
10.3C: Glycosidic bond hydrolysis
Acetals can be hydrolyzed back to hemiacetals. Notice that an acetal to hemiacetal
conversion is an SN1-type reaction with a water nucleophile and an alcohol leaving group.
Mechanism for acetal hydrolysis (enzyme-catalyzed):
R1
C
OR2
H
OR2
R1 C OR3
R1
step 1
H
acetal
H A
C
OR2
OR2
H
O
+
H
HOR3
step 2
H
:B
R1 C OH
H
hemiacetal
fig 20
In step 1, an alcohol is protonated by a nearby acid group as it breaks away to form a
resonance-stabilized carbocation intermediate. The carbocation is attacked by a
nucleophilic water molecule in step 2 to form a hemiacetal.
The general mechanism above applies to reactions catalyzed by glycosidase enzymes,
which catalyze the cleavage of glycosidic bonds in carbohydrates. In the introduction to
this chapter, we learned about ongoing research in the field of cellulosic ethanol. Recall
that the main bottleneck in the production of ethanol from sources such as switchgrass or
wood is the cellulase-catalyzed step in which the glycosidic bonds in cellulose are
cleaved. Cellulose-digesting microbes have several different but closely related forms of
cellulase enzymes, all working in concert to cleave cellulose into smaller and smaller
pieces until individual glucose molecules are free to be converted to ethanol by the
fermentation process. Below is a representative mechanism for a cellulase reaction.
64
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Cellulase mechanism:
enz
O
H O
OH
cellulose
O
O
HO
OH
cellulose
O
O
HO
H
OH
O
H
OH
OH
cellulose
O
O
HO
+
HO
OH
OH
O
resonance-stabilized
carbocation intermediate
H
H
O
OH
O
O
cellulose
O
HO
H
H
enz
OH
cellulose
O
HO
OH
O
cellulose
OH
H
O
HO
O
H
OH
OH
fig 21
The starch-digesting amylase enzymes used in the corn ethanol production process
catalyze similar glycoside hydrolysis reactions, the main difference being the opposite
stereochemistry at the anomeric carbon of the substrate.
Exercise 10.8: Notice that the glycososide bond-forming reaction requires the cell to
'spend' a high-energy UTP molecule, but the cellulase reaction does not. Think in terms
of energetics - what is the fundamental difference between the two reactions that could
explain this observation?
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
65
Chapter 10: Nucleophilic addition to carbonyls
Exercise 10.9: Below is the structure of the artificial sweetener sucralose. Identify the
two anomeric carbons in the disaccharide.
HO
OH
HO
Cl
Cl
O
HO
O
O
OH
Cl
sucralose
Exercise 10.10: Robinose is a disaccharide found in 'Chenille Plant', a flowering shrub
native to the Pacific Islands.
CH3
OH
O
HO
O
HO
O
OH
OH
OH
OH
robinose
fig 21c
a) Identify the two anomeric carbons and the glycosidic bond in robinose.
b) Based on what you know of glycosidic bond-forming reactions in nature, propose a
reasonable mechanism for the linking of the two monosaccharides, starting with the
activated hemiacetal (as you don't know the identity of the phosphate group, use 'X' as in
figure 18).
c) Based on what you know of glycosidase reactions, draw a likely mechanism for the
enzymatic hydrolysis of the glycosidic bond in robinose.
d) Using the same carbon numbering system as for glucose in figure x.x, fill in in the
carbon numbers (#1 through #6) for each of the monosaccharides that make up robinose.
e) Draw the open chain form of each of the monosaccharides
Exercise 10.11: Look again at the structures of the two-glucose fragments of cellulose
and amylose shown the introduction to this chapter. A structural feature of the cellulose
polymer makes it inherently more resistant to enzymatic hydrolysis compared to starch.
Explain. (Hint: think about intermolecular interactions.)
66
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Section 10.4: N-glycosidic bonds
We have just seen that when a second alcohol attacks a hemiacetal or hemiketal, the
result is an acetal or ketal, with the glycosidic bonds in carbohydrates providing a
biochemical example. But if a hemiacetal is attacked not by a second alcohol but by an
amine, what results is a kind of ‘mixed acetal’ in which the anomeric carbon is bonded to
one oxygen and one nitrogen.
R
N
RO C
R
H
R
fig 22a
This arrangement is referred to by biochemists as an N-glycosidic bond. You may
recognize these as the bonds in nucleosides and nucleotides that link the G, C, A, T, or U
base to the sugar.
NH2
N
N
PPPO
O
N
N
H
OH OH
ATP
N-glycosidic bond
R
N
RO C
R
H
R
fig 22
The formation of N-glycosidic bonds in ribonucleotides is closely analogous to the
formation of glycosidic bonds in carbohydrates – again, it is an SN1-like process with an
activated water leaving group. Typically, the hemiacetal is activated by
diphosphorylation, as illustrated in step A of the general mechanism below.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
67
Chapter 10: Nucleophilic addition to carbonyls
Mechanism for formation of an N-glycosidic bond:
H
ATP
AMP
OH
RO C
H
R
hemiacetal
step A
(see section 9.x)
PPi
OPP
RO C
R
H
step 1
N
R
R
R
RO C
H
R
activated hemiacetal
R
step 2
resonance-stabilized
carbocation intermediate
RO C
N H
RO C
H
R
N-glycoside
H
R
fig 22b
The starting point for the biosynthesis of purine (G and A) ribonucleotides is a fivecarbon sugar called ribose-5-phosphate, which in solution takes the form of a cyclic
hemiacetal. The critical N-glycosidic bond is established through substitution of NH3 for
OH at the anomeric carbon of the ribose. The anomeric OH group is first activated (step
A below) to form an activated intermediate called phosphoribosylpyrophosphate (PRPP).
The inorganic pyrophosphate then leaves to generate a resonance-stabilized carbocation
(step 1) which is attacked by a nucleophilic ammonia in step 2 to establish the Nglycosidic bond.
68
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
O
PO
OH
hemiacetal
HO
OH
ribose-5-phosphate
ATP
step A
(see section x.x)
N-glycosidic bond
AMP
O
PO
OPP
NH3
O
PPi
PO
NH3
step 2
step 1
HO
O
PO
HO
OH
HO
OH
OH
PRPP
many steps
O
PO
purine ribonucleotides
HO
OH
fig 23
With the N-glycosidic bond in place, the rest of the purine base is assembled piece by
piece by other biosynthetic enzymes.
(The mechanism in figure 23 should look familiar - we saw step A in chapter 9 as an
example of alcohol diphosphorylation , and steps 1 and 2 in chapter 8 as an example of a
biochemical SN1 reaction).
Establishment of the N-glycosidic bond in biosynthesis of the pyrimidine ribonucleotides
and (U, C and T) also begins with PRPP, but here the ring structure of the nucleotide base
part of the biomolecule has already been 'pre-fabricated' in the form of orotate:
H
O
O
PO
H
OPP
PPi
O
N
PO
+
O
HO
OH
PRPP
O
N
CO2
N
O
N
O2C
H
HO
OH
orotate
many steps
pyrimidine ribonucleotides
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
69
Chapter 10: Nucleophilic addition to carbonyls
fig 23a
Exercise 10.12: We have just seen an illustration of the formation of an N-glycosidic
bond in a biosynthetic pathway. In the catabolic (degradative) direction, an N-glycosidic
bond must be broken, in a process which is analogous to the hydrolysis of a glycosidic
bond (illustrated earlier). In the catabolism of guanosine nucleoside, the N-glycosidic
bond is broken by inorganic phosphate (not water!) in an SN1-like displacement reaction.
Predict the products of this reaction, and draw a likely mechanism. Assume that the
phosphate attacks from the bottom side of the sugar ring, as drawn.
O
N
N
H
HO
O
N
NH2
N
Pi
H
OH OH
guanosine
Exercise 10.13: Glycoproteins are proteins that are linked, by glycosidic or N-glycosidic
bonds, to sugars or carbohydrates through an asparagine, serine, or threonine side chain
on the protein. As in other glycosylation and N-glycosylation reactions, the hemiacetal of
the sugar must be activated prior to glycosidic bond formation. Below is the structure of
the activated sugar hemiacetal substrate in an asparagine glycosylation reaction. (Nature
2011, 474, 350)
O O
O O
P
O
O P
O
O
R
HO
O
H
N
O
OH
protein
CH3
H2N
asparagine side chain
OH
activated sugar
fig 21a
Draw a likely mechanism for the asparagine glycosylation reaction. Hint: remember,
these reactions occur at the anomeric carbon of the sugar - start by finding it! Assume
that the configuration of the anomeric carbon is inverted in the course of the reaction.
70
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Section 10.5: Imines and iminium ions
The electrophilic carbon atom of aldehydes and ketones can be the target of nucleophilic
attack by amines as well as alcohols. The end result of attack by an amine nucleophile is
a functional group in which the C=O double bond is replaced by a C=N double bond, and
is known as an imine. (An equivalent term is 'Schiff base', but we will use 'imine'
throughout this book). Recall from section 7.5B that imines have a pKa of approximately
7, so at physiological pH they can be accurately drawn as either protonated (iminium ion
form) or neutral (imine).
Iminium ion formation
R
C
H2O
RNH3
O
R
C
R
pKa ~ 7
N
R
R
H
iminium ion
Mechanism (enzymatic):
H A
R
H A
O
OH
C
R C R
R
step 1
R
N H
H
R
R
step 2
:B
C
R
+
H2O
N
R
H
NH2
fig 25
Mechanistically, the formation of an imine involves two steps. First, the amine nitrogen
attacks the carbonyl carbon in a nucleophilic addition step (step 1) which is closely
analogous to hemiacetal and hemiketal formation. Based on your knowledge of the
mechanism of acetal and ketal formation, you might expect that the next step would be
attack by a second amine to form a compound with a carbon bound to two amine groups
– the nitrogen version of a ketal or acetal. Instead, what happens next (step 2 above) is
that the nitrogen lone pair electrons ‘push’ the oxygen off of the carbon, forming a C=N
double bond (an iminium) and a displaced water molecule.
The conversion of an iminium back to an aldehyde or ketone is a hydrolytic process
(bonds are broken by a water molecule), and mechanistically is simply the reverse of
iminiom formation:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
71
Chapter 10: Nucleophilic addition to carbonyls
Hydrolysis of an iminium ion:
R
R
C
H2O
R
RNH3
NH
O
R
C
R
Mechanism (enzymatic):
H
:B
H
O
R
C
R
:B
H
O
R
R
C
O
R
R
NH2
NH
R
C
+
R
RNH3
H A
H A
fig 26
Carbon-carbon bond forming enzymes called aldolases (which we'll cover in detail in
chapter 12) often form iminium links between a carbonyl carbon on a substrate and a
lysine residue from the active site of the enzyme, as in this aldolase reaction from the
Calvin Cycle:
enzyme
H3N
+
enzyme
O
OP
H2O
OH
HN
OP
OH
fig 27
After the carbon-carbon bond forming part of an aldolase reaction is completed, the
iminium linkage is hydrolyzed, freeing the product so that it can diffuse out of the active
site and allow another catalytic cycle to begin.
In chapter 17, we will learn about reactions that are dependent upon a coenzyme called
pyridoxal phosphate (PLP), also known as vitamin B6. In these reactions, the aldehyde
carbon of PLP links to an enzymatic lysine in the active site:
72
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
enzyme
NH3
+
enzyme
H
O
OH
PO
N
N
H
H2O
OH
PO
CH3
N
CH3
fig 27c
Then, the PLP-lysine imine linkage is traded for an imine linkage between PLP and the
amino group on the substrate, in what can be referred to as a transimination.
enzyme
N
H
H
OH
PO
N
R
+
amino acid substrate
PLP-lysine
O
R
H
O
NH3
CH3
H
O
O
N
OH
PO
N
+
enzyme
NH3
CH3
PLP-amino acid
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
73
Chapter 10: Nucleophilic addition to carbonyls
The mechanism for a transimination is very similar to that of imine formation:
Tansimination:
R
H
C
R
NH
H
RNH3
+
C
R
NH
RNH3
+
R
Mechanism:
H A
R
NH2
H
C
H
B:
R
R
NH
HN C
R
H A
H
R
NH2
H R
B:
R
H
C
NH
+
RNH3
R
fig 27e
Exercise 10.14:
a) Propose a mechanism for the following reaction. (Hint - first, identify the imine in the
product).
OH
O
OH
H2O
NH3
O
O
O
O
O
O
N
O
O
fig 27a
b) Draw a likely mechanism and predict the product of this iminium hydrolysis step from
the proline degradation pathway.
H
H2O
N
CO2
74
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Section 10.6: A look ahead: addition of carbon and hydride nucleophiles to
carbonyls
We have seen in this chapter a number of reactions in which oxygen and nitrogen
nucleophiles add to carbonyl groups. Other nucleophiles are possible in carbonyl
addition mechanisms: in chapters 12 and 13, for example, we will examine in detail some
enzyme-catalyzed reactions where the attacking nucleophile is a resonance stabilized
carbanion (usually an enolate ion):
H A
O
O
R
C
R
C
R
C
O
H OH
C
C
R
R
C
R
H
H
H
nucleophile is a resonancestabilized carbanion
O
R
C
C
R
H
fig 28
Then in chapter 15, we will see how the carbonyl groups on aldehydes and ketones can
be converted to alcohols through the nucleophilic addition of what is essentially a hydride
(H-) ion.
H A
O
H
R
C
OH
R
R C R
H
fig 29
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
75
Chapter 10: Nucleophilic addition to carbonyls
Key concepts for review
Before moving on to the next chapter, you should be confident in your ability to:
Recognize aldehyde and ketone groups in organic biomolecules
Draw/explain the bonding picture for aldehyde and ketone groups
Explain why the carbonyl carbon in an aldehyde or ketone is electrophilic
Draw complete curved arrow mechanisms for the following reaction types:
formation of a hemiacetal/hemiketal
collapse of a hemiacetal/hemiketal to revert to an aldehyde/ketone
formation and hydrolysis of an acetal/ketal
formation and hydrolysis of an N-glycosidic bond
formation and hydrolysis of an imine
transimination
Explain how the carbocation intermediates in glycosidic bond formation and hydrolysis
reactions are stabilized by resonance
Explain the stereochemical considerations of a nucleophilic addition to an
aldehyde/ketone, especially in the context of glycosidic bond formation. Be able to
identify the re and si faces of an aldehyde, ketone, or imine.
In addition to these fundamental skills, you should develop your confidence in working
with end-of-chapter problems involving more challenging, multi-step biochemical
reactions.
76
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
Problems
P10.1 Draw a mechanism showing the formation of an imine linkage between a lysine
side chain and a-ketobutyrate (this is the first step in the degradation of lysine).
O
O
O
O
O
α-ketoglutarate
P10.2: Draw four possible cyclic hemiketal isomers of the compound below.
O
O2C
HO
OH
OP
OH
P10.3: A downstream intermediate in the lysine degradation pathway undergoes imine
hydrolysis to release two amino acid products. Draw a mechanism for this hydrolysis
reaction, and show the structures of the two products formed.
O
O
O
H2O
NH3
O
N
O
O
P10.4: Below is the structure of lactose, the sugar found in dairy products.
OH
OH
O
HO
OH
HO
H
O
OH
H
OH
O
OH
lactose
Lactose is a disaccharide of galactose and glucose. People who are lactose intolerant do
not produce enough lactase - the enzyme that hydrolyzes the glycosidic bond linking the
two monosaccharides - to be able to fully digest dairy products.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
77
Chapter 10: Nucleophilic addition to carbonyls
a) Draw a likely stabilized carbocation intermediate in the hydrolysis reaction catalyzed
by lactase.
b) Draw, in the chair conformation, the structure of what you predict would be the most
abundant form of the galactose monosaccharide in aqueous solution.
c) Is galactose an aldose or a ketose?
d) Draw, showing sterochemistry, the open-chain form of galactose.
P10.5: You probably know that ascorbic acid (vitamin C) acts as an antioxidant in the
body. When vitamin C does its job, it ends up being oxidized to dehydroascobate, which
is usually drawn as shown below, in the so-called tricarbonyl form.
O
OH
O
O
O
dehydroascorbate
(tricarbonyl form)
Evidence suggests, however, that the most important form of dehydroascorbate in a
physiological context is one in which two of the ketone groups are in their hydrated form,
and one is an intramolecular hemiketal (see Chemical and Engineering News, Aug. 25,
2008, p. 36). Show the structure of this form of dehydroascorbic acid.
78
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
P10.6: The compound below is the product of a ring-opening imine hydrolysis step in the
degradation pathway for proline, one of the amino acids. Draw the structure of the
starting compound.
NH3
CO2
O
P10.7: The rearrangement below was proposed to proceed via imine formation followed
by nucleophilic substitution. Propose a mechanism that fits this description. (J. Biol.
Chem. 280, 12858, scheme 2 part 2) .
R
O
R
O
H2N
HO
O
HN
N
N
H
H
O
O
P10.8: The biochemical acetal-forming reactions we learned about in this chapter all
require activation of the hemiacetal through phosphorylation. In the organic synthesis
lab, non-enzymatic acetal-forming reactions are carried out with a catalytic amount of
strong acid, which serves to activate the hemiacetal. Predict the product of the following
acetal forming reaction, and propose a reasonable mechanism for the reaction.
Remember that the reaction is carried out under acidic conditions, which means that the
protonation state of intermediates will be different than biochemical reactions occurring
at neutral pH.
O
+
H+
HO
OH
Problems 9-15 all involve variations on, and combinations of, the nucleophilic addition
steps that we studied in this chapter. Although the reactants and/or products may look
somewhat different from the simpler aldehydes, acetals, imines, etc. that we used as
examples in the chapter, the key steps still involve essentially the same mechanistic
patterns. Before attempting these problems, you may want to review tautomerization
reactions in section 7.6.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
79
Chapter 10: Nucleophilic addition to carbonyls
P10.9: The final step in the biosynthesis of inosine monophosphate (IMP, a precursor to
both AMP and GMP), is a ring-closing reaction in which a new nitrogen-carbon bond
(indicated by an arrow in the structure below) is formed. Predict the starting substrate for
this reaction, and propose a mechanism that involves a slight variation on typical imine
formation.
O
N
H2O
PO
O
N
N
H
N
?
HO
OH
inosine monophosphate
P10.10: Propose a mechanism for these steps in nucleotide metabolism:
a)
NH2
H2O
O
NH4
HN
N
N
O
N
O
R
R
Cytidine
Uridine
b)
NH2
N
N
N
N
R
Adenosine
80
H2O
NH4
O
N
N
NH
N
R
Inosine
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 10: Nucleophilic addition to carbonyls
P10.11
a) Draw the structure (including stereochemistry) of the compound that results when the
cyclic hemiketal shown below coverts to an open-chain compound with two ketone
groups.
HO CO2
O
H2C
OH
OH
b) The compound shown below undergoes a ring-opening reaction to form a species that
can be described as both an enol and an enamine. Draw the structure (including
stereochemistry) of this product, and a likely mechanism for its formation.
OP
O
HO
H
N
CO2
OH
P10.12: Tetrahydrofolate (THF) is a coenzyme that serves as a single-carbon donor in
many biochemical reactions. Unlike S-adenosylmethionine (SAM, see section 8.8), the
carbon being transferred in a THF-dependent reaction is often part of a carbonyl. Below
is a reaction in the histidine degradation pathway. The mechanism involved is thought to
be an transimination, followed by a imine-to-imine tautomerization, followed by an imine
hydrolysis. Propose a reasonable mechanism that fits this description. Hint: first identify
the carbon atom being transferred.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
81
Chapter 10: Nucleophilic addition to carbonyls
H2O
NH4
NH3
NH
HN
O2C
O2C
CO2
CO2
glutamate
+
+
R
H
N
R
N
H
HN
O
R
HN
R
N
R
N
H
R
5-formyl THF
THF
P10.13: Hydrazones are close relatives of imines, formed in reactions between
aldehydes/ketones and hydrazines, a functional group containing a nitrogen-nitrogen
bond. The mechanism for hydrazone formation is analogous to that of imine formation.
Guanafuracin, a known antibiotic compound, is a hydrazone, and can be prepared easily
in the laboratory by combining equimolar amounts of the appropriate aldehyde and
hydrazine in water (no heat or acid catalyst is required, and the reaction is complete in
seconds).
Determine the starting materials required for the synthesis of guanafuracin, and propose a
likely mechanism for the reaction.
R
H2O
O
R
C
+
H
aldehyde
N
R
H2N N
R
H
hydrazine
C
N
H
hydrazone
NH
N
O2N
H
N
H
guanafuracin
82
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
NH2
Chapter 10: Nucleophilic addition to carbonyls
P10.14: Propose reasonable mechanisms for the following steps from the histidine
biosynthesis pathway, and predict the structure of intermediate A (which is open-chain,
not cyclic).
R1
R1
N
N
N
H2N
N
NH3
N
NH
O
O
+
intermediate A
NH2
H2N
O
R2
H2O
NH
N
R2
P10.15: (Challenging!) Propose a likely mechanism for the synthesis of glucosamine 6phosphate from fructose-6-phosphate. One of several intermediates is shown. (EC
2.6.1.16.) Arch. Biochem. Biophys. 2007 474, 302, scheme 4).
OP
PO
O
OH
O
H2O
NH3
OH
HO
NH3
HO
OH
OH
OH
glucosamine-6-phosphate
fructose-6-phosphate
enz
PO
OH
HO
N
OH
OH
intermediate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
83
Chapter 11: Nucleophilic acyl substitution
Chapter 11
Nucleophilic acyl substitution reactions
Introduction
The 26th of July, Notice is given to the Sheriffs, that in the Street of Lescalle, a Part of the old
Town inhabited only by poor People, Fifteen Persons are suddenly fallen sick: They dispatch
thither Physicians and Surgeons; they examine into the Distemper, and make Report; some, that
'tis a Malignant Fever; others, a contagious or pestilential Fever, occasioned by bad Food, which
Want had long forced those poor Creatures to live upon . . .
The 27th, Eight of those Sick dye; the Sheriffs themselves go to their Houses to cause them to be
searched; Buboes [swelling of the lymph nodes] are found on Two of them: The Physicians and
Surgeons still hold the same Language, and impute the Cause of the Distemper to unwholsome
Food. Notwithstanding which, as soon as Night comes, M. Moustier repairs to the Place, sends for
Servants from the Infirmaries, makes them willingly or by Force, take up the Bodies, with all due
Precautions; they are carried to the Infirmaries, where they are buried with Lime; and all the rest
of the Night he causes the remaining Sick, and all those of their Houses, to be removed to the
Infirmaries.
84
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
The 28th, very early in the Morning, Search is made every where for those who had
Communication with them, in order to confine them: Other Persons in the same Street fall sick,
and some of those who first sicken'd dye. ..
The People who love to deceive themselves, and will have it absolutely not to be the Plague, urge
a Hundred false Reasons on that Side. Would the Plague, say they, attack none but such poor
People? Would it operate so slowly?
Let them have but a few Days Patience, and they will see all attacked without Distinction, with the
swiftest Rage, and the most dreadful Havock, that ever was heard of.
(source: Gutenberg Project
http://www.gutenberg.org/files/45673/45673-h/45673-h.htm)
In late May of 1720, a ship arrived in the Mediterranean port city of Marseille, having
recently departed from Cyprus and Tripoli. Although several crew members had fallen ill
and died during the journey, the ship was allowed to unload after only a very brief
quarantine, the result of political pressure on port authorities from local businessmen who
wanted quick access to the valuable silk and cotton waiting in the ship's hold.
Along with silk and cotton, the hold carried rats. The rats, in turn, carried fleas. The fleas
carried a microscopic mass murderer: Yersinia pestis, the species of bacteria that causes
bubonic plague.
It is next to impossible to estimate how many people have died from bubonic plague over
the course of human history. In the time of the 'Black Death' in the 14th century, it wiped
out more than half the population of Europe. In the Great Plague of Marseille in 1720,
over 100,000 people succumbed to Y. pestis infection in the city and surrounding
provinces. At the height of the outbreak, corpses piled up in city streets, and a fortified
wall, the 'mur de la peste' was constructed in an attempt to prevent people from traveling
north to the neighboring city of Aix.
Throughout history, bacteria have been the cause of untold human death and suffering,
making the threat posed by more obviously frightening species - lions and bears, spiders
and snakes – seem inconsequential by comparison. As recently as the mid-1940s, a minor
cut or cold could become a life-threatening event if a bacterial infection were to set in,
and even in developed countries, one in twenty infants did not survive to celebrate their
first birthday.
Since then, the infant mortality rate in developed countries has declined by a factor of
ten. You probably don't worry very much when a small cut on your hand becomes
infected. The idea of half of the population of the United States dying in a plague is, in
most people's minds, the stuff of zombie movies, not reality. Bacteria are, for now at
least, no longer public enemy #1.
How did this happen?
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
85
Chapter 11: Nucleophilic acyl substitution
For an answer, we move to a September morning in 1928, in the laboratory of Alexander
Fleming, a Scottish bacteriologist working at St. Mary's Hospital in London. As a young
man serving in the British Medical Corps during World War I, Fleming saw first-hand
how deadly bacteria could be, as he watched countless soldiers in his battlefield hospital
die from infected wounds. After returning to civilian life, he began to study
Staphylococci bacteria, a common source of life-threatening infections in humans, hoping
to discover new antibacterial agents that were more effective than those he had used in
the war. He spent a lot of his time growing Staphylococcus cultures in petri dishes for his
experiments, and, notoriously untidy, he tended to leave piles of culture dishes lying
around his lab. One morning, he returned from a short vacation to find that one of the
cultures he had left out had some mold growing on it. He was about to throw it away, but
happened to notice something curious: surrounding the small spot of mold was a circle of
clear medium, where no bacteria were growing. He realized that the mold must be
secreting something that killed bacteria.
As it turned out, the mold was a of strain called Penicillium notatum, and the 'something'
killing the bacteria was an organic compound that came to be known as penicillin.
Ph
O
H
HN
S
C
O
N
CH3
CH3
CO2
penicillin
Fleming published his findings in the British Journal of Experimental Pathology, but
made only passing reference to the potential therapeutic value of penicillin. The paper
received little attention.
Fast-forward now to early February 1941, with the world once again at war. One
morning, a policeman named Albert Alexander living in Oxford, England, had an
unfortunate gardening accident. While he was trimming some roses on his day off, his
shears slipped and gave him a nasty cut on the side of his mouth. The cut became
infected, and after a few days it appeared as if the infection would kill him. Then, he got
a visit in his hospital room from some chemists at nearby Oxford University.
For the last few years, the chemists had been hard at work isolating pure penicillin from
mold cultures, a tricky job because the compound tends to degrade during purification. It
is a feat that Alexander Fleming -who, after all, was a bacteriologist, not a chemist - had
never been able to accomplish, but the Oxford researchers had realized how valuable
penicillin might be to the war effort, and had finally met with some success. They needed
a human subject on whom to test the ability of their compound to treat infected wounds,
and Albert was their man for the job. They injected him with penicillin, and within a day
his infection cleared up. It was a new day in the history of medicine.
86
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
At the heart of a penicillin molecule is an amide functional group - more specifically, a
cyclic amide, or 'lactam'. To understand how penicillin works at the molecular level as it
prevents bacteria from multiplying, we first need to know more about the chemistry of
amides and other carboxylic acid derivative functional groups, and a type of organic
reaction mechanism called 'nucleophilic acyl substitution'. Understanding the reactivity
of carboxylic acid derivative groups will also allow us to appreciate why penicillin is so
prone to degradation, and why - very significantly for all of us - the era of not having to
worry about bacterial infections may be near an end, as common toxic bacterial species
such as Staphylococcus develop increasingly robust resistance to antibiotics.
Section 11.1: Carboxylic acid derivatives
The functional groups at the heart of this chapter are called carboxylic acid derivatives:
they include carboxylic acids themselves, carboxylates (deprotonated carboxylic acids),
amides, esters, thioesters, and acyl phosphates.
O
R
C
OH
R
carboxylic acid
R
C
O
S
R
O
carboxylate
O
C
O
O
R
C
C
R
R
O
O
R
C
C
N
R
amide R
O
O P O
ribose-A
O
O
thioester
R
ester
O
O P O
O
acyl-AMP
acyl phosphate
fig 1
Cyclic esters and amides are referred to as lactones and lactams, respectively.
O
O
O
N
a lactone
a lactam
H
fig 2a
Carboxylic acid anyhydrides and acid chlorides, which also fall under the carboxylic acid
derivative category, are not generally found in biomolecules but are useful intermediates
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
87
Chapter 11: Nucleophilic acyl substitution
in laboratory synthesis. They are discussed in a section on laboratory reactions at the end
of this chapter.
O
O
R
C
C
R
Cl
acid chloride
O
C
O
R
acid anhydride
fig 2
Carboxylic acid derivatives can be distinguished from aldehydes and ketones by the
presence of a group containing an electronegative heteroatom - usually oxygen, nitrogen,
or sulfur – bonded directly to the carbonyl carbon. You can think of a carboxylic acid
derivative as having two sides. One side is the acyl group, which is the carbonyl plus the
attached alkyl (R) group. In the specific cases where R is a hydrogen or methyl, chemists
use the terms formyl and acetyl group, respectively. One the other side is the
heteroatom-linked group: in this text, we will sometimes refer to this component as the
‘acyl X' group (this, however, is not a standard term in organic chemistry).
'acyl X' group
acyl group
formyl group
acetyl group
O
R
O
C
X
R
H
C
O
X
R
H3C
C
X
R
fig 3
Notice that the acyl X groups are simply deprotonated forms of other functional groups
linked to the acyl group: in an amide, for example, the acyl X group is an amine, while in
an ester the acyl X group is an alcohol.
O
H
R
acyl group
=
R
H
O
C
N
R
C
N
H
amide
amine
plus
O
H
O
R
alcohol
=
R
C
ester
ig 4
Exercise 11.1: What is the ‘acyl X’ group in:
a) an acid anhydride?
b) a thioester?
c) a carboxylic acid?
d) an acyl phosphate?
88
R
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
O
R
Chapter 11: Nucleophilic acyl substitution
Exercise 11.2: Draw the structures indicated:
a) compound A after is has been acetylated (ie. an acetyl group added)
b) compound B after it has been formylated
c) compound C after it has been formylated
d) Compound D after it has been acetylated
O
H
O
A
O
CH3
H3C N
O
OH
D
OH
B
SH
C
'Fatty acid' molecules such as stearate are carboxylates with long carbon chains for acyl
groups.
O
O
stearate
fig 5
The aromas of many fruits come from small ester-containing molecules:
H3C
O
pear
O
O
O
H
O
O
O
CH3
apple
raspberry
O
pineapple
fig 6
The 'peptide bonds' that link amino acids together in proteins are amides.
'peptide bond'
(amide)
O
protein
CH3
N
CH3
protein
H
fig 7
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
89
Chapter 11: Nucleophilic acyl substitution
Acetyl-Coenzyme A, a very important two carbon (acetyl group) 'building block'
molecule in metabolism, is characterized by reactions at its thioester functional group:
NH2
thioester
N
O
H3C
S
O
O
N
N
H
H
O
O P O P O
O
OH
N
O
N
O
N
O
O
acetyl-CoA
OH
O P O
O
fig 8
Exercise 11.3: There are two amide groups in acetyl-CoA: identify them.
Exercise 11. 4: Name any/all carboxylic acid derivative groups in the molecules below
(one or more molecules may not have any!)
H
O
O
O
CH3
O
O
CH3
O
S
OH
H2N
O
O
O
N
H3N
H
O
O
O
H3N
HO
O
HO
O
CO2
HO
CH3
OH
OH
O
O
CH3
fig 8a
Section 11.2: The nucleophilic acyl substitution mechanism
The fact that one of the atoms adjacent to the carbonyl carbon in carboxylic acid
derivatives is an electronegative heteroatom – rather than a carbon like in ketones or a
hydrogen like in aldehydes - is critical to understanding the reactivity of carboxylic acid
derivatives. The most significant difference between a ketone/aldehyde and a carboxylic
90
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
acid derivative is that the latter has a potential leaving group - what we are calling the
'acyl X group' - bonded to the carbonyl carbon.
ketone
(or aldehyde)
carboxylic acid derivative
(eg. thioester)
R
C
X
R
H3C
C
O
O
O
O
S
CH3
R
C
H3C
R
leaving group
C
CH2CH3
not a leaving group
fig 9a
As a result, carboxylic acid derivatives undergo nucleophilic acyl substitution reactions,
rather than nucleophilic additions like ketones and aldehydes.
A nucleophilic acyl substitution reaction starts with nucleophilic attack at the carbonyl,
leading to a tetrahedral intermediate (step 1 below). In step 2, the tetrahedral
intermediate collapses and the acyl X group is expelled, usually accepting a proton from
an enzymatic acid in the process.
Mechanism for a nucleophilic acyl substitution reaction:
O
B:
H Nu
R
C
O
X
step 1
R
O
H A
R C X
Nu
R
step 2
C
R
Nu
+
H X
R
fig 9
Notice that in the product, the nucleophile becomes the new acyl X group. This is why
this reaction type is called a nucleophilic acyl substitution: one acyl X group is
substituted for another. For example, in the reaction below, one alcohol 'X group'
(methanol), substitutes for by another alcohol 'X group' (3-methyl-1-butanol) as one ester
is converted to another.
acetyl group
O
O
H3C
C
O
+ CH3OH
H3C
C
OCH3
methanol
+
HO
3-methyl-1-butanol
fig 10
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
91
Chapter 11: Nucleophilic acyl substitution
Another way of looking at this reaction is to picture the acyl group being transferred from
one acyl X group to another: in the example above, the acetyl group (in green) is
transferred from 3-methyl-1-butanol (blue) to methanol (red). For this reason,
nucleophilic acyl substitutions are also commonly referred to as acyl transfer reactions.
Enzymes catalyzing nucleophilic acyl substitution reactions have evolved ways to
stabilize the negatively charged, tetrahedral intermediate, thus lowering the activation
energy of the first, rate-determining step (nucleophilic attack). The late transition state of
the first step resembles the tetrahedral intermediate that results: recall from chapter 6 that
the Hammond postulate tells us that anything that stabilizes the tetrahedral intermediate
will also stabilize the transition state. In many cases, for example, enzymatic amino acid
residues are positioned in the active site so as to provide stabilizing hydrogen bond
donating interactions with the negatively-charged oxygen. This arrangement is sometimes
referred to in the biochemistry literature as an oxanion hole. The figure below shows a
tetrahedral intermediate stabilized by hydrogen bond donation from two main chain
(amide) nitrogen atoms.
enz
enz
O
enz
N
N
H
O
enz
H
O
R C XR
Nu
fig 9b
Section 11.3: The relative reactivity of carboxylic acid derivatives
In carboxylic acid derivatives, the partial positive charge on the carbonyl carbon is
stabilized by electron donation from nonbonding electrons on the adjacent heteroatom,
which has the effect of decreasing electrophilicity.
O
O
ketones / aldehydes
R
C
R
R
R
C
R
O
O
carboxylic acid
derivative
(eg. thioester)
C
S
R
R
C
O
S
R
R
C
S
R
lone pair on 'acyl X' group makes
carbonyl carbon less electropositive
92
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
fig 11
Among the carboxylic acid derivatives, carboxylate groups are the least reactive towards
nucleophilic acyl substitution, followed by amides, then carboxylic esters and carboxylic
acids, thioesters, and finally acyl phosphates, which are the most reactive among the
biologically relevant acyl groups. Acid anhydrides and acid chlorides are laboratory
reagents that are analogous to thioesters and acyl phosphates, in the sense that they too
are highly reactive carboxylic acid derivatives. Section 11.8 near the end of this chapters
includes information about the chemistry of these two reagents.
Relative reactivity of carboxylic acid derivatives:
O
O
R
C
SR
thioester
O
O
R
C
R
O
C
O
N
H
carboxylate
amide
R
R
C
R
carboxylic acid
C
O
O P O
O
acyl phosphate
(or acyl-AMP)
O
OH
R
C
OR
ester
O
O
O
C
C
R
R
O
acid anhydride
R
C
Cl
acid chloride
most reactive
least reactive
fig 12
The reactivity trend of the carboxylic acid derivatives can be understood by evaluating
the basicity of the leaving group (acyl X group) - remember from section 8.4 that weaker
bases are better leaving groups. A thioester is more reactive than an ester, for example,
because a thiolate (RS-) is a weaker base and better leaving group than an alcoxide (RO-).
Recall from chapter 7 that the pKa of a thiol is about 10, while the pKa of an alcohol is
15 or higher: a stronger conjugate acid means a weaker conjugate base.
In general, if the incoming nucleophile is a weaker base than the ‘acyl X’ group that is
already there, it will also be the better leaving group, and thus the first nucleophilic step
will simply reverse itself and we’ll get the starting materials back:
O
O
R
C
B:
X
R
R
R C
X
Nu
Nu
H A
H
back where we started!
O
R
C
better leaving group
X
R
Nu
H
fig 13
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
93
Chapter 11: Nucleophilic acyl substitution
In general, acyl substitution reactions convert higher energy carboxylic acid derivatives
into derivatives of lower energy. Thioesters, for example, are often converted directly
into carboxylic esters in biochemical reactions, but not the other way around. To go
'uphill' - from a carboxylate to a thioester, for example, requires the 'coupling' of the
uphill reaction to an energetically favorable reaction. We will see how this works in the
next section.
Section 11.4: Acyl phosphates
Acyl phosphates, because they are so reactive towards acyl substitutions, are generally
seen as reaction intermediates rather than stable metabolites in biochemical pathways.
Acyl phosphates usually take one of two forms: a simple acyl monophosphate, or acyladenosine monophosphate.
NH2
O
R
C
O
O P O
N
O
R
C
O P O
N
O
N
O
O
HO
acyl monophosphate
N
O
OH
acyl adenosine monophosphate
(acyl AMP)
fig 14
Both forms are highly reactive to acyl substitution reactions, and are often referred to as
‘activated acyl groups’ or ‘activated carboxylic acids’ for reasons that will become
clear soon. The tendency of phosphates to form stabilizing complexes with one or more
magnesium ions in an enzyme's active site contributes in a large way to the reactivity of
acyl phosphates.
Mg+2
Mg+2
O
R
C
O
O
O P O
O
R
C
O
coordination to Mg2+ draws electron
density away from acyl carbon,
making it more electrophilic
O P O
O
very electrophilic carbon
fig 15
A magnesium ion acts as a Lewis acid, accepting electron density from the oxygen end of
the acyl carbonyl bond, which greatly increases the degree of partial positive charge and thus the electrophilicity - of the carbonyl carbon. The magnesium ion also balances
negative charge on the phosphate, making it a weak base and excellent leaving group.
94
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
We have already learned that the carboxylate functional group is the least reactive
substrate for an enzyme-catalyzed acyl substitution reactions. In biology, though,
carboxylates are frequently transformed into thioesters, carboxylic esters, and amides, all
of which are higher in energy, meaning that these transformations are thermodynamically
'uphill'.
O
R
C
S
R
thioester
O
R
C
R
O
carboxylic ester
O
R
C
N
R
H
amide
'uphill' substitutions!
O
R
C
O
carboxylate
fig 16
How are these uphill substitutions accomplished? They are not carried out directly:
like all thermodynamically unfavorable reactions in biochemistry, they are linked to an
energy-releasing, 'downhill' reaction. In this case, (and many others), the linked reaction
that 'pays for' the uphill reaction is hydrolysis of ATP.
In order to undergo an acyl substitution reaction, a carboxylate must first be activated by
phosphorylation. You are already familiar with this phosphoryl group transfer process
from chapter 9.
In many cases, enzymes activate a carboxylate group by converting it to an acyl
phosphate (the most reactive of the carboxylic acid derivatives), at the expense of an
ATP: the mechanism for this type of transformation is shown in Section 9.5.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
95
Chapter 11: Nucleophilic acyl substitution
Formation of an acyl phosphate (see section 9.5 for the complete mechanism):
ATP
ADP
O
R
C
O
R
O
O
C
carboxylate has been activated
O P O
O
carboxylate
acyl phosphate
abbreviation:
O
R
C
OP
fig 18a
As a common alternative, some enzymatic reactions begin with the conversion of a
carboxylate to an acyl-AMP intermediate:
Formation of an acyl -AMP (see section 9.5 for the complete mechanism):
ATP
PP i
O
O
R
C
O
R
C
O
O P O
ribose-A
O
carboxylate
acyl-AMP
abbreviation:
O
R
C
OAMP
fig 19a
In either case, once the carboxylate group has been activated, the reactive acyl
phosphate/acyl-AMP intermediate can go on to act as the electrophile in an energetically
favorable nucleophilic acyl substitution reaction.
96
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
O
R
C
O
O P O
acyl phosphate readily
undergoes acyl substitution
acyl phosphate O
O
O P O
O
PPi
O
activated by phosphorylation
R
ATP
X
O
carboxylate is not reactive
in acyl-substitution
R
C
O
C
XR
amide
ester
thioester
carboxylate
fig 19b
You have probably heard ATP referred to as the 'energy currency' molecule. The
reactions in this section provide a more concrete illustration of that concept. A lowerenergy group (a carboxylate) is converted to a higher-energy group (a thioester, for
example) by 'spending' a high-energy ATP.
Section 11.5: Formation of thioesters, carboxylic esters, and amides
11.5A: Thioester formation
Thioesters, which are themselves quite reactive in acyl substitution reactions (but less so
than acyl phosphates), play a crucial role in the metabolism of fatty acids The ‘acyl X
group’ in a thioester is a thiol.
Coenzyme A is a thiol-containing coenzyme that plays a key role in metabolism.
Coenzyme A is often abbreviated 'HSCoA' in order to emphasize the importance of the
thiol functionality.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
97
Chapter 11: Nucleophilic acyl substitution
NH2
thiol
HS
H
H
N
N
O
N
OH
O
O P O P O
O
O
N
O
N
O
N
O
O
Coenzyme A
(HSCoA)
OH
O P O
O
fig 22
Coenzyme A serves as a 'carrier' group in lipid biosynthesis, and is attached by a thioester
linkage to growing fatty acid chains. Palmityl is shown below as an example of a typical
fatty acyl-CoA thioester.
NH2
O
C
acyl group
S
H
H
N
N
O
N
OH
O
O P O P O
O
O
thioester
N
O
O
N
N
O
HO
palmityl-CoA
OH
abbreviations:
O
R
C
O
SCoA
fatty acyl-CoA
R
C
OH
fatty acid
fig 23
As we look at reactions involving thioesters in this and future sections, we will frequently
see Coenzyme A playing a key role. We will also see the formation and breaking of
thioester linkages between an acyl group and other thiol-containing species, such as a
cysteine residue on the enzyme:
98
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
reaction substrate
enz
O
R
C
S
cysteine residue in
active site of enzyme
fig 24
The term 'thioesterification' refers to the formation of a thioester functional group. In a
typical biochemical thioesterification reaction, a carboxylate is first converted into an
acyl phosphate (in other words, it is activated), then the acyl phosphate undergoes an acyl
substitution reaction with a thiol nucleophile.
Thioesterification reaction:
ATP
O
R
C
ADP
Pi
+
O
HS R
C
R
O
carboxylate
S
R
thioester
Mechanism:
activation phase:
ATP
O
R
C
O
carboxylate
ADP
(see section x.x)
O
R
C
O
O P O
O
acyl phosphate
acyl substitution phase:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
99
Chapter 11: Nucleophilic acyl substitution
O
R
C
O
O
R C
O P O
S
H
O P O
S
O
R
O
O
R
O
C
S
R
thioester
R
+
:B
O
O P O
O
inorganic phosphate
(Pi)
fig 25
Fatty acids such as palmitate , from fats and oils in your food, are converted to a
coenzyme A thioester prior to being broken down by the fatty acid degradation pathway.
O
C
O
palmitate
ATP
activation of carboxylate
ADP
O
C
OP
palmityl phosphate
HSCoA
thioesterification
Pi
O
C
SCoA
palmityl CoA
fig 27
A transthioesterification reaction is a thioester to thioester conversion - in other words,
an acyl group is transferred from one thiol to another.
100
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Transthioesterification:
O
O
C
R
S
R1
+
HS
R2
R
C
S
R2
+
HS
R1
Mechanism:
O
R
C
O
S
R1
R C
S
S
S
R2
R2
H
O
R1
R
C
H A
R2
+
HS
:B
S
R1
fig 28
For example, when your body synthesizes fatty acids, the two-carbon fatty acid 'building
block' acetyl CoA is first converted to acetyl ACP (EC 2.3.1.38). ACP is an abbreviation
for 'Acyl Carrier Protein', a modified protein with a thiol-containing prosthetic group
attached to one of its serine side chains. Throughout the fatty acid chain elongation
process, the growing hydrocarbon chain remains linked to ACP.
O
H3C
O
SCoA
H3C
SACP
acetyl ACP
acetyl CoA
SACP =
O
S
O
N
N
H
H
O
O P O
OH
O
acyl carrier protein
fig 29
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
101
Chapter 11: Nucleophilic acyl substitution
Exercise 11.5: The pyruvate dehydrogenase complex (EC 1.2.4.1) catalyzes one of the
most central of all central metabolism reactions, the conversion of pyruvate to acetylCoA, which links the gycolytic pathway to the citric acid (Krebs) cycle. The reaction is
quite complex, and we are not yet equipped to follow it through from start to finish (we
will finally be ready to do this in section 17.3). The final step, however, we can
understand: it is a transthioesterification, involving a dithiol coenzyme called
dihydrolipoamide and coenzyme A. Given the information below, draw out a reasonable
mechanism for the reaction.
dihydrolipoamide
CoA
O
H3C
SH
acetyl-CoA
R
S
fig 31
Exercise 11.6: Ubiquitin is a protein which plays a key role in many cellular processes by
reversibly attaching to other proteins, thus altering or regulating their function. Recently,
a team of researchers uncovered details of the mechanism by which ubiquitin
(abbreviated Ub) is transferred by the ubiquitin activating enzyme (abbreviated E1) to
target proteins. In the first part of this process, the carboxy terminus of ubiquitin is
linked to a cysteine side chain on E1, as shown in the incomplete reaction sequence
below. Complete the figure by drawing the structures of species A and B.
cysteine on E1 enzyme
C-terminus of ubiquitin
Ub
H
O
N
C
ATP
O
SH
PPi
E1
AMP
A
B
O
fig 31b
11.5B: Formation of esters
Esterification refers to the formation of a new ester functional group.
In a typical biochemical esterification, a thioester is subjected to nucleophilic attack from
an alcohol, leading to the formation of an ester and a thiol.
102
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Esterification reaction (from thioester)
O
R
C
O
R
S
+
HO
R
R
C
O
R
+
R
HS
Mechanism:
O
R
C
O
S
R
R C
S
O
O
R
R
H
O
R
R
C
H A
O
R
+
HS
:B
R
fig 32
The reaction below is from the synthesis of triacylglycerol, the form in which fat is stored
in our bodies.
Phase 1 (transthioesterification):
R
O
O
C
C
R
SCoA
enzyme
HSCoA
+
S
fatty acyl-CoA
enzyme
HS
Phase 2 (esterification):
OH
O
R
C
S
O
HO
enzyme
+
R
O
R
C
OH
O
monoacylglycerol
R
O
O
diacylglycerol
O
enzyme
HS
fig 33
The reaction, catalyzed by monoacylglycerol acyltransferase (EC 2.3.1.22), begins (phase
1 above) with a preliminary transthioesterification step in which the fatty acyl group is
transferred from coenzyme A to a cysteine residue in the active site of the enzyme. Recall
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
103
Chapter 11: Nucleophilic acyl substitution
that it is a common strategy for enzymes to first form a covalent link to one substrate
before catalyzing the 'main' chemical reaction.
In phase 2 of the reaction, the fatty acyl group is now ready to be transferred to glycerol,
trading its thioester linkage to the cysteine for a new ester linkage to one of the alcohol
groups on glycerol.
An esterification reaction has tremendous importance in the history of drug development,
a story that we heard in the introduction to this chapter. The discovery of penicillin was
arguably one of the most important events in the history of modern medicine. The key
functional group in penicillin is the four-membered lactam (recall that a lactam is a cyclic
amide).
Ph
O
H
HN
S
lactam
C
O
N
CH3
CH3
CO2
benzyl penicillin
fig 42
Penicillin, and later generations of antibiotic drugs, have saved countless lives from oncedeadly bacterial infections. The elucidation of the chemical mechanism of penicillin
action was also a milestone in our developing understanding of how drugs function on a
molecular level. We now know that penicillin, and closely related drugs such as
ampicillin and amoxycillin, work by inhibiting an enzyme that is involved in the
construction of the peptide component bacterial cell walls. The details of the wallbuilding reaction itself are outside the scope of this discussion, but it is enough to know
that the process involves the participation of a nucleophilic serine residue in the active
site of the enzyme. The penicillin molecule is able to enter the active site, and once
inside, the lactam group serves as an electrophilic 'bait' for the nucleophilic serine:
104
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Ph
H
HN
O
serine
B:
H
Ph
O
S
C
O
N
O
H
HN
S
CH3
CH3
O
serine
CO2
C
O
CH3
N
CH3
CO2
H A
Ph
bacterial enzyme has been
'tagged' with a bulky group
serine
O
HN
O C
O
H
S
HN
CH3
CH3
CO2
4-membered lactam has been
opened, ring strain relieved
fig 43
Although you might expect that an amide-to-ester conversion such as what is shown
above would be energetically unfavorable based on the reactivity trends we have learned,
this lactam is in fact much more reactive than an ordinary amide group due to the effect
of ring strain: recall from section 3.2 that four-membered rings are highly strained, and
considerable energy is released when they are opened.
Ring strain also accounts for why penicillin has a tendency to degrade: when in contact
with water, the lactam will spontaneously hydrolyze over time, which opens the ring and
forms a carboxylate group.
Unfortunately, many strains of bacteria have acquired an enzyme called β-lactamase (EC
3.5.2.6), that catalyzes rapid hydrolysis of the lactam ring in penicillin-based drugs,
rendering them inactive. These bacteria are consequently resistant to penicillin and
related antibiotics. As you are probably aware, the evolution of drug resistance in
bacteria is a major, world-wide health problem, and scientists are engaged in a constant
battle to develop new antibiotics as the older ones become less and less effective.
In a transesterification reaction, one ester is converted into another by an acyl
substitution reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
105
Chapter 11: Nucleophilic acyl substitution
Mechanism for a transesterification reaction:
O
R
C
O
O
R
R C
O
O
O
R
R
H
O
R
C
R
H A
O
R
+
HO
:B
R
fig 34
If studying organic chemistry sometimes gives you a headache, you might want to turn to
a transesterification reaction for help. Prostaglandins are a family of molecules that
promote a wide range of biological processes, including inflammation. Acetylsalicylic
acid, commonly known as aspirin, acts by transferring - through a transesterification
reaction - an acetyl group to a serine residue on the enzyme responsible for the
biosynthesis of prostaglandin H2 (one member of the prostaglandin family).
enzyme has been 'acetylated':
now it is inactive
O
O
H3C
C
+
O
CO2
enzyme
HO
H3C
C
enzyme
O
+
HO
serine residue on
prostaglandin H2 synthase
CO2
fig 35
Acetylation of this serine blocks a channel leading to the active site, effectively shutting
down the enzyme, impeding prostaglandin production, and inhibiting the inflammation
process that causes headaches.
In section 11.8, we will see two laboratory acyl substitution reactions that lead to the
formation of aspirin and ibuprofen.
Exercise 11.7: Discuss the key structural feature of aspirin that makes it so effective at
transferring its acetyl group - in other words, why is the ester group in aspirin more
reactive than a typical ester?
106
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
11.5C: Amide formation
An activated carboxylate group (in other words, acyl phosphate or acyl-AMP) can be
converted to an amide through nucleophilic attack by an amine.
Mechanism for amide formation:
carboxylate group has been
activated by phosphoryolation
(see section 11.4 for mechanism)
R = H or ribose-A
O
R
C
O
R
O P OR
O
H 2N
B:
H
O
O
C
O P O
N
H
R
O
R
C
O
R
N
R
H
amide
+
O
O P OR
O
Pi or AMP
fig 18
The amino acid biosynthesis pathways provide examples of amide formation in biology.
The amino acid glutamine is synthesized in most species by converting the carboxylate
side chain of glutamate (another amino acid) to an amide, after first activating the
carboxylate by monophosphorylation: (EC 6.3.1.2)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
107
Chapter 11: Nucleophilic acyl substitution
H3N
C
ADP
ATP
O
O
O
H3N
activation phase
C
O
O P O
CO2
CO2
O
NH3
glutamate
acyl substitution
Pi
O
H3N
C
NH2
CO2
glutamine
fig 20
A similar process takes place in the synthesis of asparagine from aspartate, except that
the activated carboxylate in this case is an acyl-AMP:
NH3
O2C
ATP
O
C
O
PPi
activation phase
NH3
O
C
O2C
O
O P O
ribose-A
O
aspartate
NH3*
acyl substitution
AMP
NH3
O2C
O
C
NH2
asparagine
fig 21
*In the asparagine synthesis reaction, the ammonia nucleophile actually comes from hydrolysis of a
glutamine molecule.
Exercise 11.8: A enzyme in bacteria is thought to be responsible for resistance to a class
of antibiotics that includes apramycin, ribostamycin and paromomycin. The enzyme
catalyzes acetylation of the antibiotic compound by acetyl-CoA. The structure of
acetylated apramycin is shown below.
108
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
OH
O
H2N
HO
CH3
HN
HO
OH
O
H3C
O
O
HN
O
acetylated apramycin
HO
O
NH2
HO
fig 21a
a) Identify the acetyl group that has been transferred to apramycin, (and thus inactivating
it).
b) This reaction represents the conversion between what two carboxylic acid derivative
functional groups?
c) Thinking about it from a difference perspective, what functional group on the
antibiotic is the acyl acceptor?
d) Draw a reasonable mechanism for the acyl substitution reaction that occurs in this
transformation. Use abbreviations as appropriate.
Section 11.6: Hydrolysis of thioesters, esters, and amides
So far we have been looking at the formation of thioesters, carboxylic esters, and amides,
starting from carboxylates. In hydrolytic acyl substitution reactions, nucleophilic water is
the incoming nucleophile and a carboxylate group is the final product. Because
carboxylates are the least reactive among the carboxylic acid derivatives, these hydrolysis
reactions are thermodynamically favorable, with thioester hydrolysis the most favorable
of the three.
Thioester, carboxylic ester, and amide hydrolysis:
O
R
C
S
C
O
C
H2O
ROH
R
O
R
RSH
R
O
R
H2O
R
H2O
N
O
C
O
RNH2
R
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
109
Chapter 11: Nucleophilic acyl substitution
Mechanism:
X=
S (thioester)
O (carboxylic ester)
N(H) (amide)
O
R
C
O
R
X
R C
X
O
H
H A
H
O
+
:B
:B
C
R
O
H
O
R
HX
R
thiol, alcohol, or amine
fig 37
In the citric acid (Krebs) cycle, (S)-citryl CoA is hydrolyzed to citrate (EC 2.3.3.8):
HO CO2 O
O2C
C
H2O
HSCoA
HO CO2 O
O2C
C
SCoA
O
citrate
(S)-citryl-SCoA
fig 38
Acetylcholinesterase (EC 3.1.1.7), an enzyme present in the synapse, catalyzes
hydrolysis of the ester group in acetylcholine, a neurotransmitter that triggers muscle
contraction.
O
H3C
C
H2O
N
O
CH3
H3C
acetylcholine
O
CH3
H3C
C
O
+
acetate
N
HO
H3C
CH3
CH3
choline
fig 3915)
Like many other hydrolytic enzymes, the acetylcholinesterase reaction proceeds in two
phases: first, a covalent enzyme-substrate intermediate is formed when the acyl group of
acetylcholine is transferred to an active-site serine on the enzyme (a transesterification
110
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
reaction). A water nucleophile then attacks this ester, driving off acetate and completing
the hydrolysis.
Exercise 11.9: Based on the above description, draw the structure of the covalent
enzyme-substrate intermediate in the acetylcholinesterase reaction.
If the action of acetylcholinesterase is inhibited, acetylcholine in the synapse does not get
hydrolyzed and thus accumulates, resulting in paralysis and death in severe cases. Sarin
nerve gas is a potent inhibitor of acetylcholinasterase action. Some victims of the Tokyo
subway sarin attack in 1995 who were exposed to low levels of the gas reported that they
initially realized that something was wrong when they noticed how dark everything
seemed around them. This was due to uncontrolled contraction of their pupils. You will
be invited to consider the mechanism of inhibition by sarin in problem x.x
Peptide (amide) bonds in proteins and polypeptides are subject to spontaneous
(nonenzymatic) hydrolysis in water.
amide group
(peptide bond)
protein
C
R
H2O
R
O
N
H
R
O
protein
C
protein
O
R
+
H
N
protein
H
fig 40
Although this amide to carboxylate conversion is thermodynamically a downhill reaction,
peptide bonds are kinetically very stable (they react slowly) at neutral pH. In fact, the
half-life for uncatalyzed hydrolysis of a peptide bond in pH 7 water is by some estimates
as high as 1000 years. (Ann. Rev. Biochem. 2011, 80, 645.)
The stability of peptides bonds makes good physiological sense: we would all be in
trouble if our enzymes, receptors, and structural proteins were hydrolyzing away while
we slept! That being said, it is also true that controlled, specific hydrolysis of peptide
bonds, catalyzed by a large, diverse class of enzymes called proteases, is a critical
biochemical reaction type that can occur very rapidly, in many different biological
contexts. For example, many proteins only become active after they have been
‘processed' - in other words, hydrolyzed at a specific amino acid location by a specific
protease.
Although all proteases catalyze essentially the same reaction – amide hydrolysis different protease subfamilies have evolved different catalytic strategies to accomplish
the same result. HIV protease is the target of some the most recently-developed antiOrganic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
111
Chapter 11: Nucleophilic acyl substitution
HIV drugs. It plays a critical role in the life cycle if the HIV virus, hydrolyzing specific
peptide bonds of essential viral proteins in order to convert them to their active forms.
HIV protease is a member of the aspartyl protease subfamily, so-named because of the
two aspartate residues located in the active sites of these enzymes. HIV protease is also,
as you are probably aware, the target of HIV protease inhibitor drugs, which are a
component of the most effective treatment currently available for HIV infection.
In HIV protease and other aspartyl proteases, the two enzymatic aspartates residues
(shaded grey and abbreviated 'Asp1' and 'Asp2 'in the figure below) work in concert to
activate the electrophile, nucleophile, and leaving group in the reaction.
O
O
Asp1
Asp1
O
O
H
H
R
O
protein
C
R
H
O
N
OH R
O
protein
C
protein
step 1
H
R
N
protein
H
O
O
H
H
Asp2
O
Asp2
O
step 2
O
protein
C
R
O
O
R
protein
C
O
H
+
H2N
protein
R
+
R
H3N
protein
fig 41 Exactly how this works is a subject of some debate and the details may well vary
according to the enzyme in question, but one likely mechanism is illustrated in the figure
above, where Asp1, which is initially in its protonated form, contributes a hydrogen bond
to draw electron density away from the carbonyl carbon, making it more electrophilic. At
the same time, Asp2, which begins the reaction cycle in its anionic form, deprotonates the
water molecule to make it more nucelophilic. In step 2, Asp2 donates a proton back to
the nitrogen, making it a better leaving group.
112
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
HIV protease inhibitors shut down this reaction, which prevents the virus from
processing the proteins that it uses to bond to host cells.
Exercise 11.10: Lactonase (EC 3.1.1.17), the second enzyme in the oxidative branch of
the pentose phosphate pathway, catalyzes hydrolysis of the lactone (cyclic ester) group in
6-phosphogluconolactone. Draw the structure of 6-phosphogluconate, the product of
this reaction, and propose a reasonable mechanism for the reaction. For the mechanism,
assume direct hydrolysis, without formation of a covalent enzyme-substrate intermediate.
OP
H2O
O
HO
HO
HO
6-phosphogluconate
O
6-phosphogluconolactonate
fig 41a
Exercise 11.11: Draw the product of the β-lactamase-catalyzed hydrolysis of penicillin as
described in section 11.6.
Exercise 11.12: Draw the structure of the missing product (designated below by question
marks) in the reaction below, which is part of degradation pathway for the amino acid
tryptophan. How could you describe this reaction in organic chemistry terminology?
NH3
O
H2O
???
O2C
H
O
NH3
O
O2C
N
H2N
H
fig 41b
Section 11.7: Protein synthesis on the ribosome
Recall from section 1.3D that the 'peptide bonds' which link amino acids to form
polypeptides and proteins are in fact amide functional groups. The figure below shows
the first four amino acid residues in a protein, starting at the amino terminus.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
113
Chapter 11: Nucleophilic acyl substitution
peptide bonds (amides) are circled
N-terminus
R2
O
H3N
H
R1
aa1
C-terminus
N
N
H
N
R3
O
aa2
R4
O
aa3
H
O
aa4
fig 56
Let’s take a look at the chemistry behind the formation of a new peptide bond between
the first two amino acids - which we will call aa-1 and aa-2 - in a growing protein
molecule. This process takes place on the ribosome, which is essentially a large
biochemical 'factory' in the cell, composed up of many enzymes and RNA molecules, and
dedicated to the assembly of proteins. You will learn more in a biochemistry or cell
biology course about the complex but fascinating process of ribosomal protein synthesis.
For now, we will concentrate on the enzyme-catalyzed organic transformation that is
taking place: the formation of an amide from a carboxylate and an amine.
We have seen amide-forming reactions before – think back to the glutamine and
asparagine synthetase reactions (section 11.5C). The same ideas that we learned for
those reactions hold true for peptide bond formation: the carboxylate group on a substrate
amino acid must first be activated, and the energy for this activation comes from ATP.
The carboxylate group of aa-1 is first transformed to an acyl-AMP intermediate through a
nucleophilic substitution reaction at the α-phosphate of ATP.
ATP
O
H3N
O
R1
PPi
O
H3N
O
O
R1
ribose-A
P O
O
activated acyl-AMP intermediate
fig 57
In the next step, the amino acid is transferred to a special kind of RNA polymer called
transfer RNA, or tRNA for short. We need not concern ourselves here with the structure
of tRNA molecules- all we need to know for now is that the nucleophile in this reaction is
a hydroxyl group on the terminal adenosine of a tRNA molecule. Because this tRNA
molecule is specific to aa-1, we will call it tRNA-1
114
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
NH2
tRNA
N
N
O
O P O
N
O
N
abbreviation:
O
tRNA
HO
OH
OH
fig 58
The incoming nucleophile is an alcohol, thus what we are seeing is an esterification: an
acyl substitution reaction between the activated carboxylate of aa-1 and an alcohol on
tRNA-1 to form an ester.
aa1 is now linked to its tRNA
O
ribose-A
O
H3N
H3N
O P O
R1
O
AMP
O
O
R1
+
tRNA-1
OH
tRNA-1
fig 59
This reaction, starting with activation of the amino acid, is catalyzed by a class of
enzymes called aminoacyl-tRNA synthetases (there are many such enzymes in the cell,
each one recognizing its own amino acid - tRNA pair).
The first amino acid is now linked via an ester group to tRNA-1. The actual peptide
bond-forming reaction occurs when a second amino acid (aa-2) also linked to its own
tRNA-2 molecule, is positioned next to the first amino acid on the ribosome. In another
acyl substitution reaction, catalyzed by an enzymatic component of the ribosome called
peptidyl transferase (EC 2.3.2.12), the amino group on aa-2 displaces tRNA1: thus, an
ester has been converted to an amide (thermodynamically downhill, so ATP is not
required).
new peptide bond
O
H3N
R2
O
R1
aa-1
+
O
H3N
O
tRNA-1
aa-2
R2
O
tRNA-2
H3N
tRNA-2
O
N
R1
H
O
+
OH
tRNA-1
dipeptide
fig 60
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
115
Chapter 11: Nucleophilic acyl substitution
This process continues on the ribosome, as one amino acid after another is added to the
growing protein chain:
R2
O
H3N
tRNA-2
H3N
O
N
R1
O
+
H
R3
O
dipeptide
H3N
H
O
N
N
R1
H
tRNA-3
aa-3
R2
O
O
O
R3
tRNA-2
+
O
OH
tRNA-3
tripeptide
fig 61
When a genetically-coded signal indicates that the chain is complete, an ester hydrolysis
reaction – as opposed to another amide formation - occurs on the last amino acid, which
we will call aa-n. This reaction is catalyzed by proteins called release factors (RFs).
C-terminus
H
Rn
O
N
tRNA-n
H
H
O
N
Rn-1
H2O
protein
O
Rn
O
N
O
N
Rn-1
H
O
+
tRNA-n
OH
fig 62
This hydrolysis event frees the mature protein from the ribosome, and results in the
formation of a free carboxylate group at the end of the protein (this is called the carboxyterminus, or C-terminus of the protein, while the other end – the ‘starting’ end – is called
the N-terminus).
116
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Section 11.8: Nucleophilic acyl substitution reactions in the lab
All of the biological nucleophilic acyl substitution reactions we have seen so far have
counterparts in laboratory organic synthesis. Mechanistically, one of the biggest
differences between the biological and the lab versions is that the lab reactions usually
are run with a strong acid or base as a catalyst, whereas biological reactions are of course
taking place at physiological pH. When proposing mechanisms, then, care must be taken
to draw intermediates in their reasonable protonation states: for example, a hydronium
ion (H3O+) intermediate is reasonable to propose in an acidic reaction, but a hydroxide
(OH-) intermediate is not.
11.8A: Ester reactions - bananas, soap and biodiesel
Acid-catalyzed synthesis of flavor compounds such as isopentyl acetate (an ester with the
flavor of banana) is simple to carry out in the lab. In this esterification reaction, acetic
acid is combined with isopentyl alcohol along with a catalytic amount of sulfuric acid.
Acid-catalyzed esterification (laboratory reaction):
O
H3C
C
OH
acetic acid
+
O
H2SO4
HO
H3C
isopentanol
C
+
O
H2O
isopentyl acetate
Mechanism:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
117
Chapter 11: Nucleophilic acyl substitution
H
O
H3C
C
O
protonation increases
electrophilicity of carbonyl carbon
O S OH
H
H
O
H3C
O
H
O
C
OH
O
OH
step 1
H3C
C
H3C C OH
step 2
OH
R
O
H
HO
O
H
H
step 3
O
O
H3C C OH2
C
R
step 4
H3C
O
O
O S OH
R
step 5
O
+ H2O
O
HO S OH
O
O
H3C
C
O
isopentyl acetate
fig 46
The carbonyl oxygen of acetic acid is first protonated (step 1), which draws electron
density away from the carbon and increases its electrophilicity. In step 2, the alcohol
nucleophile attacks: notice that under acidic conditions, the nucleophile is not
deprotonated simultaneously as it attacks (as we would show in a biochemical
mechanism), and the tetrahedral intermediate is a cation rather than an anion. In step 3, a
proton is transferred from one oxygen atom to another, creating a good leaving group
(water) which is expelled in step 4. Finally (step 5), the carbonyl oxygen on the ester is
deprotonated, regenerating the catalytic acid.
This reaction is highly reversible, because carboxylic acids are approximately as reactive
as esters. In order to obtain good yields of the ester, an excess of acetic acid can be used,
which by Le Chatelier's principle (see your General Chemistry textbook for a review)
shifts the equilibrium toward the ester product.
118
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Saponification is a common term for base-induced hydrolysis of an ester. For example,
methyl benzoate will hydrolize to benzoate and methanol when added to water with a
catalytic amount of sodium hydroxide.
Mechanism of base-catalyzed ester hydrolysis (saponification):
CH3O
O
C
O
CH3
O
O
C O
step 1
O
OH
CH3
C
O
H
step 2
H
step 3
O
C
O
+
CH3OH
fig 47
Addition of the base provides hydroxide ion to act as a nucleophile (hydroxide is of
course a better nucleophile than water) in step 1. The tetrahedral intermediate (anionic in
this case, because the reaction conditions are basic) then collapses in step 2, and the
alkoxide (CH3O-) leaves. We are not used to seeing alkoxides or hydroxides as leaving
groups in biochemical reactions, because they are strong bases - but in a basic solution,
this is a reasonable chemical step. Step 3 is simply an acid-base reaction between the
carboxylic acid and the alkoxide. Note that this is referred to as base-induced rather than
base-catalyzed because hydroxide is not regenerated, and thus a full molar equivalent of
base must be used.
The saponification process derives its name from the ancient craft of soap-making, in
which the ester groups of triacylglycerols in animal fats are hydrolized under basic
conditions to glycerol and fatty acyl anions (see section 2.5A for a reminder of how fatty
acyl anions work as soap).
O
O
C
+
O
C
O
O
C
OH
O
R
O
OH
3
R
C
O
+
OH
fatty acid
OH
glycerol
R
R
triacylglycerol
soap
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
119
Chapter 11: Nucleophilic acyl substitution
fig 48
We learned earlier about transesterification reactions in the context of the chemical
mechanism of aspirin. Transesterification also plays a key role in a technology that is
already an important component in the overall effort to develop environmentally friendly,
renewable energy sources: biodeisel. You may have heard stories about people running
their cars on biodeisel from used french fry oil. To make biodeisel, triacylglycerols in
fats and oils can be transesterified with methanol or ethanol under basic conditions. The
fatty acyl methyl and ethyl ester products are viable motor fuels.
O
O
C
(neutralize with strong acid)
R
+
O
C
O
O
R
triacylglycerol
C
R
O
CH3O
OH
O
H+
R
C
+
OCH3
fatty acyl
methyl ester
(biodiesel)
OH
OH
glycerol
fig 36
Exercise 11.13: Give the UIPAC name for each of the fragrant esters shown in figure
11.6, and the UIPAC names for the carboxylic acid and alcohol starting materials that
could be used to synthesize the ester.
Exercise 11.14: What would happen if you tried to synthesize isopentyl acetate (banana
oil) with basic rather than acidic conditions? Would this work?
Exercise 11.15: Draw a complete mechanism for the reverse of the acid-catalyzed
esterification reaction. What would you call this reaction in organic chemistry terms?
Exercise 11.16: An alternative way to synthesize esters is to start with a carboxylic acid
and an alkyl halide.
a) Propose such a synthesis of methyl benzoate, and draw out a mechanism showing how
it would work.
b) Should you use acidic or basic conditions? Explain.
c) What limitation do these reactions have in terms of the alkyl halide that can be used?
Explain.
Exercise 11.17: Ester hydrolysis can also be carried out under acidic conditions. Draw a
complete mechanism for the acidic hydrolysis of methyl benzoate. Be careful to propose
reasonable protonation states for all intermediates!
120
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
11.8B: Acid chlorides and acid anhydrides
In the cell, acyl phosphates and thioesters are the most reactive of the carboxylic acid
derivatives. In the organic synthesis lab, their counterparts are acid chlorides and acid
anhydrides, respectively. Of the two, acid chlorides are the more reactive, because the
chloride ion is a weaker base and better leaving group than the carboxylate ion (the pKa
of HCl is -7, while that of carboxylic acids is about 4.5: remember, a stronger conjugate
acid means a weaker conjugate base).
Acid chlorides can be prepared from carboxylic acids using SOCl2:
SO2
Cl-
O
C
R
O
+
OH
Cl
S
O
Cl
R
C
Cl
fig 49
Acid anhydrides can be prepared from carboxylic acids and an acid chloride under basic
conditions:
O
R
C
O
+
Cl
O
C
R
R
O
O
C
C
O
+
Cl
R
acid anhydride
fig 50
Acetic anhydride is often used to prepare acetate esters and amides from alcohols and
amines, respectively. The synthesis of aspirin and acetaminophen are examples:
O
H3C
O
C
O
+
HO
O
+
O
H3C
C
O
O
C
O
CH3
acetate
O
acetylsalicylic acid
(aspirin)
O
O
C
CH3
acetic anhydride
+
H2N
OH
O
OH
H3C
C
N
H
+
O
O
C
CH3
acetate
acetaminophen
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
121
Chapter 11: Nucleophilic acyl substitution
fig 51
A carboxylic acid cannot be directly converted into an amide because the amine
nucleophile would simply act as a base and deprotonate the carboxylic acid:
O
R
C
O
OH
+
R
NH2
R
C
+
O
R
NH3
fig 52
Instead, the carboxylic acid is first converted to an acid chloride (in other words, the
carboxylic acid is activated), then the acid chloride is combined with an amine to make
the amide.
SO2
Cl-
O
R
C
+
O
SOCl2
O activated carboxylic acid
R
C
Cl
H
N
R
H
Cl
O
R
C
N
R
H
fig 53
This sequence of reactions is a direct parallel to the biochemical glutamine and
asparagine synthase reactions we saw earlier in the chapter (section 11.5A), except that
the activated form of carboxylic acid is an acid chloride instead of an acyl phosphate or
acyl-AMP.
Exercise 11.18: Propose a reasonable mechanism for the carboxylate to amide
transformation shown above.
Exercise 11.19: Show how you might prepare the following compound, starting with a
carboxylic acid and an amine.
O
N
122
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Exercise 11.20: The mechanism for the conversion of a carboxylic acid into an acid
chloride involves a chlorosulfite intermediate, and SO2 is a byproduct.
O
R
O
O
S
Cl
chlorisulfite
With this information, propose a complete mechanism for the transformation.
11.8C: Polyesters and polyamides
If you have ever had the misfortune of undergoing surgery or having to be stitched up
after a bad cut, it is likely that you benefited from our increasing understanding of
polymers and carboxylic ester chemistry. Polyglycolic acid is a material commonly used
to make dissolving sutures. It is a polyester - a polymer linked together by ester groups and is formed from successive acyl substitution reactions between the alcohol group on
one end of a glycolic acid monomer and the carboxylic acid group on a second:
carboxylic acid on
one monomer
O
HO
C
C
OH
ester
O
H H
H
H H
C
OH
HO
C
C
C
O
H H
O
H H
C
OH
HO
C
+
+
O
O
H
dimer
alcohol on
another monomer
O
HO
alcohol on 3rd
monomer
C
C
OH
H H
ester
O
O
+
H H
C
O
C
HO
C
C
C
OH
C
O
H
H
H H
O
O
H H
C
C
O
n
polyglycolic acid
ester
H
O
H
trimer
fig 44
The resulting polymer - in which each strand is generally several hundred to a few
thousand monomers long - is strong, flexible, and not irritating to body tissues. It is not,
however, permanent: the ester groups are reactive to gradual, spontaneous hydrolysis at
physiological pH, which means that the threads will dissolve naturally over several
weeks, eliminating the need for them to be cut out by a doctor.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
123
Chapter 11: Nucleophilic acyl substitution
Exercise 11.21: Dacron, a polyester used in clothing fiber, is made of alternating
dimethyl terephthalate and ethylene glycol monomers.
O
H3CO C
O
C OCH3
HOCH2CH2OH
ethylene glycol
dimethyl terephthalate
fig 54
a) Draw the structure of a Dacron tetramer (in other words, four monomers linked
together).
b) Water is a side product of glycolic acid polymerization. What is the equivalent side
product in Dacron production?
Exercise 11.22: Nylon 6,6 is a widely used polyamide composed of alternating
monomers. Nylon 6,6 has the structure shown below -the region within the parentheses
is the repeating unit, with 'n' indicating a large number of repeats. Identify the two
monomeric compounds used to make the polymer.
O
H
N
O
N
H
n
Nylon 66
fig 55
11.8D: The Gabriel synthesis of primary amines
The Gabriel synthesis, named after the 19th-century German chemist Siegmund Gabriel,
is a useful way to convert alkyl halides to amines and another example of SN2 and acyl
substitution steps in the laboratory. The nitrogen in the newly introduced amine group
comes from phthalimide. In the first step of the reaction, phthalimide is deprotonated by
hydroxide, then in step 2 it acts as a nucleophile to displace a halide in an SN2 reaction
(phthalimide is not a very powerful nucleophile, so this reaction works only with
unhindered primary or methyl halides).
124
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
O
O
OH-
H N
O
R
R
Br
N
step 1
O
N
step 2
O
O
phthalimide
step 3
(hydrolysis)
O
OH-
O
R
R
H N
NH2
R
O
primary amine
imide functional group
+
O
O
O
fig 55a
Step 3 is simply a pair of hydrolytic acyl substitution steps to release the primary amine,
with an aromatic dicarboxylate by-product.
Exercise 11.23: Phthalimide contains an 'imide' functional group, and has a pKa of
approximately 10. What makes the imide group so much more acidic than an amide,
which has a pKa of approximately 17?
As an alternative procedure, release of the amine in step 3 can be carried out with
hydrazine (H2NNH2) instead of hydroxide. Again, this occurs through two nucleophilic
acyl substitution reactions.
O
O
R
N
O
H2N NH2
H
R
NH2
primary amine
+
H
N
N
O
fig 55b
In 2000, chemists at MIT synthesizing a porphyrin-containing molecule introduced two
amine groups using the Gabriel synthesis with hydrazine. Porphyrins, which include the
'heme' in our red blood cells, are an important family of biomolecules with a variety of
biochemical functions.\ (J. Org. Chem. 2000, 65, 5298).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
125
Chapter 11: Nucleophilic acyl substitution
O
Br
Br
NH
H N
O
N
KOH
N
HN
H2NNH2
H
H
N
NH2
H2N
O
NH
N
N
HN
+
N
O
fig 55c
126
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Section 11.9: A look ahead - acyl substitution reactions with a carbon or hydride
nucleophile
Although we have seen many different types of nucleophilic acyl substitutions in this
chapter, we have not yet encountered a reaction in which the incoming nucleophile is a
carbanion or a hydride. Recall that in the previous chapter on aldehydes and ketones, we
also postponed discussion of nucleophilic carbonyl addition reactions in which a
carbanion or a hydride is the nucleophile. The reason for putting off these discussions is
that these topics are both important and diverse enough to warrant their own dedicated
chapters.
In the next chapter, we will see many carbonyl addition and acyl substitution reactions
where the nucleophilic species is a resonance-stabilized carbanion such as an enolate
(section 7.6B). Then in chapter 14, we will encounter nucleophilic addition and acyl
substitution reactions in which a hydride ion (H-) essentially plays the part of a
nucleophile. In these chapters we will see how nucleophilic carbanion and hydride
species are generated in a biochemical context. For now, see if you can predict the result
of the following biochemical reactions.
Exercise 11.24: predict the products of the following nucleophilic acyl substitution
reactions, both of which are part of the biosynthesis of isoprenoid compounds such as
cholesterol and lycopene:
a) acetoacetyl CoA acetyltransferase (enolate nucleophile)
O
O
H2C
SCoA
+
enzyme
H3C
S
product with new carbon-carbon bond
(Fig 24)
b) HMG-CoA reductase (the nucleophile here is not literally an isolated hydride ion,
which would be a very unlikely species in a physiological environment. We will learn in
chapter 16 what is actually going on, but for the time being, just predict the result of an
acyl substitution reaction with a "hydride ion" nucleophile.)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
127
Chapter 11: Nucleophilic acyl substitution
hydride equivalent
O HO CH3 O
+ :H
O
SCoA
product with aldehyde group
fig 45
128
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Key learning objectives for this chapter
Be able to recognize and draw examples of carboxylic acid derivative functional groups:
carboxylic acids/carboxylates
acyl phosphates (both acyl monophosphate and acyl-AMP)
thioesters
esters
amides
acid chlorides
carboxylic acid anhydrides
Know the meaning of the terms 'acyl', 'acetyl', 'formyl', 'lactone', and 'lactam'.
You need not memorize the structure of coenzyme A, but you should recognize that it
contains a key thiol group and often forms thioester linkages, particularly in fatty acid
metabolism.
Understand what happens in a nucleophilic acyl substitution (also called acyl transfer
reaction), and be able to draw mechanistic arrows for a generalized example.
Know the trends in relative reactivity for the carboxylic acid derivatives:
in a biological context (acyl phosphates and thioesters as activated acyl groups)
in a laboratory context (acid chlorides and carboxylic acid anhydrides as activated
acyl groups)
Recognize and understand the most important types of nucleophilic acyl substitution
reactions in biology:
How a carboxylate group, which is unreactive to nucleophilic acyl substitution
reactions, is activated in the cell by ATP-dependent phosphorylation to either
acyl monophosphate or acyl-AMP.
Conversion of an acyl phosphate to a thioester, a (carboxylic) ester, or an amide.
Transthioesterification, esterification, and transesterification reactions.
Conversion of a thioester or ester to an amide
Hydrolysis of a thioester, a (carboxylic) ester, or an amide to a carboxylate.
Understand the energetics of the above reactions:
Carboxylate to acyl phosphate is 'uphill' energetically, paid for by coupling to
hydrolysis of one ATP
Other conversions above are 'downhill': it is unlikely, for example, to see a direct
conversion of an amide to an ester.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
129
Chapter 11: Nucleophilic acyl substitution
(Notable exception: the lactam (cyclic amide) group in penicillin is very reactive
due to ring strain, and forms an ester with an active site serine residue in the
target protein)
You need not memorize all of the details of peptide bond formation on the ribosome, but
you should be able to follow the description in section 7 and recognize the nucleophilic
acyl substitution reactions that are occuring.
Be able to draw complete mechanisms for the following lab reactions:
acid-catalyzed esterification of a carboxylic acid
saponification (base-catalyzed hydrolysis of an ester), application to soap-making
base-catalyzed transesterification, application to biodiesel production
Understand how acid chlorides and carboxylic acid anhydrides serve as activated acyl
groups in laboratory synthesis. Be able to describe how an amide to ester conversion
could be carried out in the laboratory.
Understand how polyesters and polyamides are formed. Given the structure of a polymer
be able to identify monomer(s), and vice-versa.
Be able to recognize, predict products of, and draw mechanisms for the Gabriel synthesis
of primary amines, using either hydroxide ion or hydrazine to release the amine product.
130
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
Problems
P11.1: Here is some practice in recognizing carboxylic acid derivative functional groups
in large, complex biological molecules. There are seven amide and four ester groups in
the molecules below - see if you can find them all.
F
F
HO
NH2
O
H
H3CO
N
N
N
F
N
O
CF3
capsaicin
(the 'hot' molecule in hot peppers)
sitagliptin
(component of Janumet, a diabetes drug from Merck Pharmaceuticals)
H2N
O
N
H
N
O
H3C
O
O
NH
N
O HO
HN
OH
OH
O
OH
H3C
O
O
O
H
O O
O
O
N
CH3
O
O
HN
CH3
H3C CH3
Saquinavir
(first HIV protease inhibitor to be approved by FDA)
paclitaxel
(cancer drug Taxol from Bristol-Myers Squibb)
P11.2: (EC 6.3.4.16)
a) Predict the structures of compounds A, B, and C in the reaction pathway below.
Compound C contains an activated carboxylate functionality. Use abbreviations as
appropriate.
b) Draw a reasonable mechanism for the A to B step
ATP
ADP
NH3
O
HO
O
A
ATP
Pi
B
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
ADP
C
131
Chapter 11: Nucleophilic acyl substitution
P11.3: Imagine that acetylcholine is combined with acetylcholinesterase (section 11.6)
in a buffer made from 18O-labeled water. Where would you expect to find the 18O label
in the products?
P11.4: Predict the products of this hydrolysis reaction: (EC 3.5.1.18) (
O
H2O
CO2
N
O2C
H
H3N
CO2
P11.5: The breakdown of fat in our bodies begins with the action of lipase enzymes,
which catalyze the cleavage of fatty acids from the glycerol backbone of triacylglycerol
(see section 1.3A for a reminder of the structure of triacylglycerol). A serine residue in
the lipase active site plays a key nucleophilic role in the reaction. Draw the single
mechanistic step in which the covalent link between a fatty acid and the glyceryl
backbone is broken, using curved arrow notation and appropriate abbreviation.
P11.6: Before long-chain fatty acids are transported across the inner mitochondrial
membrane, they are temporarily transferred from Coenzyme A to a transport molecule
called carnitine, to which they are linked by an ester group.
H3C CH3 OH
N
H3C
O
O
carnitine
Draw the structure of fatty acyl carnitine (use R to denote the hydrocarbon chain of the
fatty acid)
P11.7: Below is a reaction from carbohydrate metabolism (EC 2.3.1.157). Identify the
compound designated with a question mark.
OH
OH
O
HO
HO
H
H2N
?
HSCoA
O
HO
H
HO
HN
OP
O
132
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OP
CH3
Chapter 11: Nucleophilic acyl substitution
P11.8: Propose the most likely enzymatic hydrolysis product of the substrate below (hint think about electrophilicity when considering regiochemical outcomes!) (EC 3.5.2.2) J. Biol.
Chem. 2006, 281, 13762 scheme 2)
O
N
N
H
H2O
O
H
P11.9: The coenzyme tetrahydrofolate (THF) participates in single-carbon transfer reactions.
One derivative of THF, called 10-formyl-THF (abbreviated structure shown below),
transfers a formyl group early in purine ribonucleotide biosynthesis to glycinamide
ribonucleotide.
10-formyl-THF + glycinamide ribonucleotide
THF + formylglycinamide ribonucleotide
O
O
H
H2N
H
R
N
N
N
N
H
OP
O
10-formyl-THF
N
O
N
H
O
H
N
HO
H
H
OH
formylglycinamide ribonucleotide
Draw a nucleophilic attack step for this reaction (assume that acyl transfer between the two
substrates is direct, without any covalent enzyme-substrate intermediates being formed).
P11.10: One of the key steps in the biosynthesis of purine nucleotides (guanosine and
adenosine) in archaea is shown below.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
133
Chapter 11: Nucleophilic acyl substitution
O
O
NH2
ATP
X
PO
O
N
NH2
ADP
Y
PO
NH2
O
NH
N
O
OH OH
H
OH OH
Identify the missing compounds X and Y in the figure above, and draw the structure of an
acyl phosphate intermediate.
P10.11: The reaction below is part of nucleotide biosynthesis. Predict the structures of
compound A, which contains a carboxylate group (EC 6.3.2.6).
ATP
Asp
A
ADP
Pi
O
CO2
N
N
N
H
NH2
R
CO2
P11.12: Recall from section 11.6 that acetylcholinesterase catalyzes the hydrolysis of the
ester group in acetylcholine, going through an intermediate in which the acetyl group on the
substrate is transferred to a serine on the enzyme by a transesterification reaction. The nerve
gas sarin acts by blocking this initial transesterification step: the drug enters the active site
and attaches to the active site serine. Given the structure of sarin below, propose a
mechanism for how this happens.
O
O P
F
CH3
sarin
P11.13: Propose a mechanism for the following reaction from histidine biosynthesis.
Mechanistically speaking, it is not technically an acyl substitution reaction, but is a close
cousin.
134
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 11: Nucleophilic acyl substitution
N
N
HN
R
N
H2O
O
N
R
N
N
N
NH2
R
HN
R
P11.14 (Challenging!) In the final step of the urea cycle (a phase of amino acid degradation
pathways), the amino acid arginine is hydrolyzed to urea and ornithine. Propose a reasonable
mechanism. EC 3.5.3.1
NH3
O
C
H2N
NH2
urea
O
H3N
ornithine
O
P11.15: In the biosynthetic pathway for the DNA/RNA bases uridine and cytidine, a reaction
occurs in which carbamoyl phosphate condenses with aspartate, and the resulting
intermediate cyclizes to form dihydroorotate. Propose a mechanism for this transformation.
Hint: in a very unusual step, a carboxylate group is at one point in the process directly
subjected to an acyl transfer reaction, without prior activation by phosphorylation. The
enzyme accomplishes this with the help of two bo.und zinc ions, which serve to stabilize the
negative charge on a hydroxide leaving group. (EC 3.5.2.3) (Biochemistry 2001, 40, 6989,
Scheme 2)
O
Asp
O
H2N
OP
H2O
Pi
HN
O
N
H
CO2
dihydroorotate
P11.16: Propose a reasonable mechanism for the reaction below (from lysine
biosynthesis), and fill in the missing species indicated by question marks.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
135
Chapter 11: Nucleophilic acyl substitution
O
O2C
N
CO2
+
C
O2C
tetrahydropicolinate
succinyl-CoA
?
?
O
HN
CO2
CO2
O2C
136
SCoA
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
Chapter 12
Reactions at the α-carbon, part I
Introduction
There are lots of things that can kill you in northern Australia. On land, there is the death
adder, the tiger snake and the redback spider; in the water, you'd be well advised to give
wide berth to the salt water crocodile, the stonefish, the great white shark, and of course,
the duck-billed platypus.
Credit: Alan Couch (https://www.flickr.com/photos/couchy/)
The duck-billed platypus?
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
137
Chapter 12: α-carbon part I
Consider this: in 1991, a man fishing a river in northern Queensland, Australia happened
across a platypus sitting on a log. Thinking it was injured, he picked it up. For his
trouble, he spent the next six days in a nearby hospital, suffering from two puncture
wounds in his right hand that resulted in "immediate, sustained, and devastating" pain,
against which the usual analgesic drugs were almost completely useless. His hand
"remained painful, swollen and with little movement for three weeks. Significant
functional impairment . . . persisted for three months".
***
Meanwhile, on the other side of the planet, deep in the rain forests that straddle the border
between eastern Peru and Brazil, a young man of the Matses tribe prepares himself to
receive the 'hunting magic'. He holds the end of a short wooden stick in a fire for a few
minutes, then removes it and presses the red-hot end into the skin of his chest, holding it
there for long enough for the skin to be burned. Then he scrapes the burned skin away,
and rubs into the wound a paste made from saliva mixed with secretions taken from the
skin of a giant leaf frog.
An American journalist named Peter Gorman, who reports having had the frog-skin paste
administered in the same manner during a visit to a Matses village, describes what
happens next:
Instantly my body began to heat up. In seconds I was burning from the inside . . . I
began to sweat. My blood began to race. My heart pounded. I became acutely
aware of every vein and artery in my body and could feel them opening to allow
for the fantastic pulse of my blood. My stomach cramped and I vomited violently.
I lost control of my bodily functions. . . (and) fell to the ground. Then,
unexpectedly, I found myself growling and moving about on all fours. I felt as
though animals were passing through me, trying to express themselves through my
body.
After the immediate violent effects pass, the Matses hunter is carried by his friends to a
hammock to recover. After sleeping for a day, he awakens to find himself with what his
people call the 'hunting magic': a state of heightened awareness, possessed of tremendous
energy and an abnormally keen sense of vision, hearing and smell. In the words of Mr.
Gorman, "everything about me felt larger than life, and my body felt immensely strong...
[I was] beginning to feel quite godlike".
***
There is a connection between the killer platypus in Australia and the 'hunting magic' in
the Amazon, and it has to do with the structure and reactivity of what organic chemists
refer to as the α-carbon: the carbon atom positioned adjacent to a carbonyl or imine
group in an organic molecule:
138
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
H CH3 H
N
C
N
C
O
R
C
C
C
R
H
O
H H H H
α-carbons on a ketone
the α-carbon of an alanine amino acid residue in a protein
fig 1a
It is this chemistry that we are going to be studying for the next two chapters. But first,
let's go back to that river in northern Australia and the fisherman who apparently didn't
pay enough attention in his high school wildlife biology class.
The platypus, along with a few species of shrews and moles, is an example of a very rare
phenomenon in nature: a venomous mammal. The male platypus possesses a pair of sharp
spurs on each of his hind legs near the ankle. These spurs are hollow, and connected by a
duct to a venom-producing gland in the thigh. The consensus among scientists who study
the platypus is that males use their venomous barbs mainly when fighting each other over
territory during mating season. Because healthy animals are often found with multiple
scars from spur wounds, a platypus who gets spurred during a fight with a rival will not
always die, but the experience is unpleasant enough that he will start looking for real
estate a healthy distance down the river.
It is not easy to milk the venom from an angry, thrashing platypus, but there are scientists
out there who have done it. It turns out that, like snake and spider venom, the venom
from a platypus spur consists of a mixture of neuroactive peptides (peptides are very
short proteins - less than 50 amino acids long). Recently, a team of biochemists from the
Universities of Sydney, Queensland, and Adelaide reported that they were able to isolate
from platypus venom two forms of a 39-amino acid peptide. Further analysis using NMR
and mass spectrometry revealed that the two forms of the peptide differed in structure
only at a single amino acid: the leucine at the #2 position. In one form, the leucine had
the L configuration (or S if using the R/S system), just like the amino acids in virtually all
other peptides and proteins found in nature. In the other form, this leucine had the
unusual D, or R configuration.
O
H3N S
H3C H
N
H
O
H
H
pep
N S
S
O
R
H
peptide with L-Leu at position 2
H3N S
H3C H
H
N
pep
N S
R
H
H
O
R
H
peptide with D-Leu at position 2
fig 1a
Peptides or proteins incorporating D-amino acids are not unheard of in nature, but this
was the first time that one had been found in a mammal. Interestingly, the venom from
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
139
Chapter 12: α-carbon part I
certain marine cone snails and spiders - and, yes, the skin of the giant leaf frog in the
Amazon rain forest - also contain neuroactive peptides with D-amino acids.
What is the advantage - to a platypus, cone snail, spider, or frog - of making a venomous
peptide with D stereochemistry on one or more of its amino acids? It all has to do with
generating diversity of shape and function. These are neuroactive peptides: each one
interacts in a very specific way with a specific neural protein, thus exerting a specific
neurological effect on the person or animal exposed to the venom. The different spatial
arrangement of atoms about the α-carbon of D- and L-amino acids will cause a peptide
with a D-leucine at position #2, for example, to fold into a different shape than its
counterpart with an L-leucine at the same position. Thus, the two peptides may bind
differently to one or more proteins in the nervous system, and ultimately may exert
different neurological effects - such as intense pain in the case of playpus venom, or the
'hunting magic' effect in the case of the peptide from frog skin. The ability to incorporate
D-amino acids greatly expands the potential structural and functional diversity of these
short peptides.
The two stereoisomeric platypus venom peptides are encoded by the same gene. The
peptides are initially synthesized using all L-amino acids, and then the leucine at position
#2 undergoes a 'post-translational modification': in other words, a specific enzyme binds
the all-L peptide after it has been synthesized on the ribosome and changes the leucine
residue to the D configuration.
It is this reaction - a stereoisomerization reaction that takes place at the α-carbon of an
amino acid - that brings us to the central topic of this chapter and the next: chemistry at
the α-carbon. The key concept to recall from what we have learned about acidity and
basicity in organic chemistry, and to keep in mind throughout this discussion, is that αprotons (in other words, protons on α-carbons) are weakly acidic. Loss of an α-proton
forms an enolate - a species in which a negative formal charge is delocalized between a
carbon and an oxygen. The 'enolate' term will be very important in the next two chapters,
because most of the reactions we see will go through an enolate intermediate.
α-carbon
O
O
R
R
H
R
R
H
an enolate
fig 1d
In this chapter, we will first see several examples of isomerization reactions, in which an
enzyme acts at the α-carbon of a substrate to catalyze the interconversion of two
constitutional isomers or stereoisomers. Then, we will be introduced to a reaction type
known as the 'aldol addition' and its reverse counterpart, the 'retro-aldol' cleavage
reaction. Up to now, we have seen plenty of reactions where bonds were formed and
140
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
broken between carbon and oxygen, nitrogen, or sulfur. Here, for the first time, we will
see how enzymes can catalyze the formation or cleavage - again, at the α position - of
carbon-carbon bonds: in other words, we will learn how an α-carbon can be either a
nucleophile or a leaving group in an enzymatic reaction. This has clear importance for an
understanding of metabolism in living things: the molecules of life, after all, are built
upon a framework of carbon-carbon bonds, and metabolism is the process by which
living cells build up and break down complex biomolecules.
It all starts with the α-carbon - and as both the Australian fisherman and the Amazonian
hunter could attest, what happens at the α-carbon can have some rather dramatic
consequences.
Section 12.1: Review of acidity at the α-carbon
Let's review what we learned in section 7.6A about the acidity of a proton on an α-carbon
and the structure of the relevant conjugate base, the enolate ion. Remember that this
acidity can be explained by the fact that the negative charge on the enolate conjugate base
is delocalized by resonance to both the α-carbon and the carbonyl oxygen.
= α-carbon
R
R
B
H H
ketone
O
O
O
R
R
R
R
H
H
enolate
fig 1b
The α-carbon on the enolate is sp2-hybridized with trigonal planar geometry, as are the
carbonyl carbon and oxygen atoms (now would be a good time to go back to sections
2.1C, 2.2B, and 2.3B to review, if necessary, the geometry of π-bonding in conjugated
systems). The pKa of a typical α-proton in aqueous solution is approximately 18-20:
acidic, but only weakly so. Recall from section 7.8, however, that the effective pKa of a
functional group on an enzyme-bound molecule can be altered dramatically by the
'microenvironment' of the active site. In order to lower the pKa of an α−proton, an
enzyme catalyzing a reaction that begins with an α-proton abstraction step must further
stabilize the negative charge that develops on the oxygen atom of the (enolate) conjugate
base. Different enzymes have evolved different strategies for accomplishing this task: in
some cases, a metal cation (often Zn+2) is bound in the active site to provide a stabilizing
ion-ion interaction. In other cases, stabilization is provided by a proton-donating group
positioned near the oxygen. As a third possibility, the active site architecture sometimes
provides one or more stabilizing hydrogen bond donor groups.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
141
Chapter 12: α-carbon part I
O
enz
O
enz
Zn+2
O
enz
O
N
H
N
H
H
enz
enz
O
O
R
R
O
R
R
H
R
R
H
H
fig 1e
In most of the mechanism illustrations in this chapter where an enolate intermediate is
depicted, stabilizing metal ions or hydrogen bond interactions will not be explicitly
drawn, for the sake of clarity. However, whenever you see an enolate intermediate in an
enzyme-catalyzed reaction, you should remember that there are stabilizing interactions in
play inside the active site.
Section 12.2: Isomerization at the α-carbon
Enolate ions, as well as enols and and enamines (section 7.6) are the key reactive
intermediates in many biochemical isomerization reactions. Isomerizations can involve
either the interconversion of constitutional isomers, in which bond connectivity is altered,
or of stereoisomers, where the stereochemical configuration is changed. Enzymes that
interconvert constitutional isomers are usually called isomerases, while those that
interconvert the configuration of a chiral carbon are usually referred to as racemases or
epimerases.
12.2A: Carbonyl isomerization
One very important family of isomerase enzymes catalyzes the shifting of a carbonyl
group in sugar molecules, often converting between a ketose and an aldose (recall that the
terms ketose and aldolse refer to sugar molecules containing ketone and aldehyde groups,
respectively).
Carbonyl isomerization:
R
142
O
H
C
O
C
H O H
C
O
R
C
H H
H
ketose
aldose
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
Mechanism:
A
H
R
O
H
C
O
C
B
H
C
R
step 1
H H
B
A
H
O
C
O
H O H
C
O
R
C
step 2
H
H
H
ene-diol
ketose
aldose
fig 8
The ketose species is first converted to its enol tautomer in step 1 (actually, this particular
intermediate is known as an 'ene-diol' rather than an enol, because there are hydroxyl
groups on both sides of the carbon-carbon double bond). Step 2 leads to the aldose, and is
simply another tautomerization step. However, because there is a hydroxyl group on the
adjacent (blue) carbon, a carbonyl can form there as well as at the red carbon.
An example is the glycolysis pathway reaction catalyzed by the enzyme triose phosphate
isomerase (EC 5.3.1.1). Here, dihydroxyacetone phosphate (DHAP) is reversibly
converted to glyceraldehyde phosphate (GAP).
new chiral center (R)
A
H
PO
H
O
C
H
C
O
PO
H
O
C
step 1
H H
:B
H O H
PO
C
O
C
O
C
step 2
H
H
:B
A
DHAP (ketose)
H
GAP (aldose)
ene-diol
fig 9
Notice that DHAP is achiral while GAP is chiral, and that a new chiral center is
introduced at the middle (red) carbon of GAP. As you should expect, the enzyme is
stereoselective: in step 2 a proton is delivered to the red carbon, from behind the plane of
the page, to yield the R enantiomer.
Also in the glycolysis pathway, glucose-6-phosphate (an aldose) and fructose-6phosphate (a ketose) are interconverted through an ene-diol intermediate (EC 5.3.1.9) by
an enzyme that is closely related to triose-phosphate isomerase.
OH
OP
OH
OH
O
OH
glucose-6-phosphate
OH
OP
OH
OH
OH
O
fructose-6-phosphate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
143
Chapter 12: α-carbon part I
fig 10
Exercise12.1: draw the ene-diol intermediate in the phosphoglucose isomerase
reaction.
12.2B: Stereoisomerization at the α-carbon
Enolates are a common intermediate in reactions where the stereochemical configuration
of a chiral α-carbon is interconverted. These are commonly referred to as racemization
or epimerization reactions, depending on whether the interconverted isomers are
enantiomers or epimers (recall that the term 'epimer' refers to a pair of diastereomers that
differ by a single chiral center).
Racemization/epimerization:
O
R1
C
C
O
R1
R3
R2 H
C
C
R3
H R2
Mechanism:
stereochemistry at αcarbon has inverted
α-carbon is now planar
O
O
R1
(R1 and R2 are different)
C
C
R2 H
R3
:B
α-proton taken
from the front
step 1
R1
C
C
O
R3
step 2
C
C
R3
H R2
R2
A
R1
H
proton delivered
from the back
fig 11
These reactions proceed though a deprotonation-reprotonation mechanism, illustrated
above. In step 1, the chiral α-carbon is deprotonated, leading to a planar, achiral enolate.
In step 2, a proton is delivered back to the α-carbon, but from the opposite side from
which the proton was taken in step 1, resulting in the opposite stereochemistry at this
carbon. Two acid-base groups, positioned at opposing sides of the enzyme's active site,
work in tandem to accomplish this feat.
The proteins and peptides in all known living things are constructed almost exclusively of
L-amino acids, but in rare cases scientists have identified peptides which incorporate Damino acids, which have the opposite stereochemistry at the α-carbon. Amino acid
144
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
racemase enzymes catalyze the interconversion of L and D amino acids. As you may
recall from the introductory section to this chapter, the venom of the male platypus
contains a neurotoxic peptide in which an L-leucine amino acid has been converted by a
racemase enzyme to D-leucine. In another example, the cell walls of bacteria are
constructed in part of peptides containing D-glutamate, converted from L-glutamate by
the enzyme glutamate racemase. (EC 5.1.1.3) (Biochemistry 2001, 40, 6199).
O2C
O2C
O2C
:B
H A
H
H3N
C
C
O
step 1
H3N
C
O
C
H
O
step 2
H3N
C
O
C
O
O
L-glutamate
D-glutamate
fig 12
A reaction (EC 5.1.3.1) in sugar metabolism involves the interconversion of the epimers
ribulose-5-phosphate and xylulose-5-phosphate. The enzyme that catalyzes this reaction
is called an 'epimerase'. (J. Mol. Biol. 2003, 326, 127).
OH
PO
5
O
OH
OH
(R)
4
2
3
PO
1
5
HO H
4
O
(S)
2
3
OH
1
HO H
xylulose-5-phosphate
ribulose-5-phosphate
fig 13
Exercise 12.2: Draw a reasonable mechanism for the ribulose-5-phosphate to xylulose-5phosphate reaction. Your mechanism should show an enolate intermediate and specify
stereochemistry throughout.
Exercise12.3: Predict the products of epimerization reactions starting with each of the
substrates shown. Hint - carbons next to imine groups can also be considered α-carbons!
OH
OH
OH
N
OH
b)
a)
O
OH
HO
OH
OH
OH
fig 13
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
145
Chapter 12: α-carbon part I
Recall from chapter 3 that a major issue with the drug thalidomide is the fact that the R
enantiomer, which is an effective sedative, rapidly isomerizes in the body to the
terotogenic (mutation-causing) S enantiomer. Note that the chiral center in thalidomide is
an α-carbon!
= α-carbon
O
O
H
O
N
N
N
N
O
spontaneous racemization
at the α-carbon
O
H
O
O
O
(S)-thalidomide
(teratogenic form)
(R)-thalidomide
(effective form)
O
O
H
N
N
O
no racemization
O
(R)-oxetano-thalidomide
fig 13a
Recently, chemists reported the synthesis of a thalidomide derivative in which the
carbonyl group is replaced by an 'oxetane' ring, with the aim of making an isotopically
stable form of the drug (because the carbonyl group has been removed, racemization is
no longer possible - there is no α-carbon!) (Org. Lett. 2013, 15, 4312.)
12.2C: Alkene regioisomerization
The position of an alkene group can also be changed through a reaction in which the first
step is abstraction of an α-proton and formation of an enolate intermediate. The
degradation pathway for unsaturated fatty acids (fatty acids whose hydrocarbon chains
contain one or more double bonds) involves the 'shuffling' of the position of a carboncarbon double bond, from a cis bond between carbon #3 and carbon #4 to a trans bond
between carbon #2 and carbon #3. This is accomplished by the enzyme enoyl CoA
isomerase (EC 5.3.3.8). (J. Biol Chem 2001, 276, 13622).
146
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
Alkene isomerization:
O
O
3
4
2
3
1
SCoA
R
4
2
R
H
1
SCoA
Mechanism:
O
3
4
2
R
H H
1
O
SCoA
A
H
O
SCoA
R
SCoA
H
R
H
:B
O
3
4
2
R
H
1
SCoA
fig 14
Exercise 12.4: Consider the structures of the substrate and product of the
isomerization reaction above. What two factors contribute to the thermodynamic
'driving force' for the transformation?
Exercise 12.5: Draw a likely mechanism for this reaction, part of the biosynthetic
pathway for menthol. Your mechanism should show an enolate intermediate.
O
O
fig 14a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
147
Chapter 12: α-carbon part I
Section 12.3: Aldol addition reactions
We arrive now at one of the most important mechanisms in metabolism: the aldol
addition.
Along with Claisen condensation reactions, which we will study in the next chapter, aldol
additions are responsible for most of the carbon-carbon bond forming events that occur in
a living cell. Because biomolecules are built upon a framework of carbon-carbon bonds,
it is difficult to overstate the importance of aldol addition and Claisen condensation
reactions in the chemistry of living things!
12.3A: Overview of the aldol addition reaction
Consider the potential pathways available to a reactive enolate intermediate once the αproton has been abstracted. We'll use acetaldehyde as a simple example. The oxygen,
which bears most of the negative charge, could act as a base, (step 2 below) and the result
would be an enol.
:B
O
H
C
C
H A
O
H
H
step 1
C
H H
C
OH
H
step 2
C
H
H
enolate
C
H
H
enol
fig 15
Alternatively, the enolate carbon, which bears a degree of negative charge, could act as a
base, which is simply the reverse of the initial deprotonation step that formed the enolate
in the first place. This of course just takes us right back to the starting aldehyde.
H A
O
H
C
O
O
C
H
H
H
C
C
H
H
H
C
H
C
H
H
enolate
fig 16
In both of these cases, the electron-poor species attacked by the enolate is an acidic
proton. What if the electron-poor species - the electrophile - is not a proton but a carbonyl
carbon? In other words, what if the enolate acts not as a base but rather as a nucleophile
in a carbonyl addition reaction? For example, the enolate of acetaldehyde could attack the
carbonyl group of a second acetaldehyde molecule. The result is the formation of a new
carbon-carbon bond:
148
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
H A
O
H
O
H
C
C
CH3
O
H
C
H
C
H
C
O HO H
C
C
H
C
CH3
H H
H
H
enolate
fig 16a
This type of reaction is called an aldol addition. It can be very helpful to think of an
aldol addition reaction as simply a nucleophilic carbonyl addition (Chapter 10) reaction
with an enolate α-carbon (rather than an alcohol oxygen or amine nitrogen) as the
nucleophile.
An aldol addition reaction:
O
O
C
H
C
R
+
H
C
O
R
H
H H
HO H
C α C
C β R
R H
β-hydroxyaldehyde
Mechanism:
H A
O
H
O
H
C
C
R
O
C
R
H H
:B
step 1
H
C
C
R
step 2
H
enolate carbon acs as
a nucleophile
O HO H
C
C
H
R
C
H R
new carbon-carbon
bond has formed
fig 17
Historically, the first examples of this mechanism type to be studied involved reactions
very similar to what is shown above: an aldehyde reacting with itself. Because the
resulting product contained both an aldehyde and an alcohol functional group, the
reaction was referred to as an 'aldol' addition, a terminology that has become standard for
reactions of this type, whether or not an aldehyde is involved. More generally, an aldol
addition is characterized as a nucleophilic addition to an aldehyde, ketone, or imine
electrophile where the nucleophile is the α-carbon in an aldehyde, ketone, imine, ester, or
thioester. The enzymes that catalyze aldol reactions are called, not surprisingly,
aldolases.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
149
Chapter 12: α-carbon part I
Note that the aldol reaction results in a product in which a hydroxide group is two
carbons away from the carbonyl, in the β position. You can think of the β-hydroxy
group as a kind of 'signature' for an aldol addition product.
Depending on the starting reactants, nonenzyatic aldol reactions can take more than one
route to form different products. For example, a reaction between acetaldehyde and 2butanone could potentially result in in three different aldol addition products, depending
on which of the three α-carbons (carbons 2, 3, and 5 below) becomes the attacking
nucleophile.
O HO
H
O
O
+
H 1
2
3
4
OH
O
OH
O
6
5
fig 18
Exercise 12.6: (Hint: for each reaction, first identify the nucleophilic and electrophilic
carbon atoms on the starting compounds!)
a) Fill in the appropriate carbon numbers for each of the three possible aldol addition
products shown above.
b) Draw arrows for the carbon-carbon bond forming step that leads to each of the three
products.
12.3B: Biochemical aldol addition reactions
Fructose 1,6-bisphosphate aldolase (EC 4.1.2.13) is an enzyme that participates in both
the glycolytic (sugar catabolism) and gluconeogenesis (sugar synthesis) biochemical
pathways. The reaction catalyzed by fructose 1,6-bisphosphate aldolase links two 3carbon sugars, glyceraldehyde-3-phosphate (GAP, the electrophile in the reaction) and
dihydroxyacetone phosphate (DHAP, the nucleophile), forming a 6-carbon product. In
the figures below, the nucleophilic and electrophilic carbons are identified with dots.
150
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
The fructose 1,6-bisphosphate aldolase reaction
β-hydroxy ketone
O
O
O
OP
OH
OP
+
OP
OH
OP
OH
GAP
DHAP
OH
OH
fructose-1,6-bisphosphate
Mechanism:
A
H
O
H
:B
O
O
H
OH
two new chiral centers
have been created
O
OH
OH
OP
H
H
OP
OP
step 1
OP
OH
step 2
OP
OH
OH
fig 19
fig 20
In step 1 of the reaction, an α-carbon on DHAP is deprotonated, leading to an enolate
intermediate. In this and many other aldolase reactions, a zinc cation (Zn2+) is positioned
in the enzyme's active site so as to interact closely with - and stabilize - the negatively
charged oxygen of the enolate intermediate. This is one important way in which the
enzyme lowers the energy barrier to the reaction.
Next, (step 2), the deprotonated α-carbon attacks the carbonyl carbon of GAP in a
nucleophilic addition reaction, leading to the fructose 1,6-bisphosphate product.
Notice that two new chiral centers are created in this reaction. This reaction, being
enzyme-catalyzed, is highly stereoselective due to the precise position of the two
substrates in the active site: only one of the four possible stereoisomeric products is
observed. The enzyme also exhibits tight control of regiochemistry: GAP and DHAP
could potentially form two other aldol products which are constitutional isomers of
fructose 1,6-bisposphate.
Exercise 12.7: Draw structures of the two other constitutional isomers that could
hypothetically form in aldol addition reactions between GAP and DHAP.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
151
Chapter 12: α-carbon part I
Along with aldehydes and ketones, esters and thioesters can also act as the nucleophilic
partners in aldol reactions. In the first step of the citric acid (Krebs) cycle, acetyl CoA (a
thioester nucleophile) adds to oxaloacetate (a ketone electrophile) (EC 2.3.3.8).
A
A
H
B:
A:
H
O
H
CoAS
H
O HO CO2
O2C
CO2
O
CoAS
oxaloacetate
CoAS
acetyl CoA
O
CO2
enol intermediate
fig 22
Notice that the nucleophilic intermediate is an enol, rather than a zinc-stabilized enolate
as was the case with the fructose 1,6-bisphosphate aldolase reaction. An enol
intermediate is often observed when the nucleophilic substrate is a thioester rather than a
ketone or aldehyde.
12.3C: Going backwards: the retro-aldol cleavage reaction
Although aldol reactions play a very important role in the formation of new carboncarbon bonds in metabolic pathways, it is important to emphasize that they can also be
reversible: in most cases, the energy level of starting compounds and products are very
close. This means that, depending on metabolic conditions, aldolases can also catalyze
retro-aldol reactions: the reverse of aldol reactions, in which carbon-carbon bonds are
broken.
A retro-aldol cleavage reaction:
O
O
O
H OH
C
C
H
R
C
R H
C
H
C
+
R
H
C
R
H H
β-hydroxycarbonyl
Mechanism:
O
:B
H
O
O H
C
C
H
C
CH3
R H
step 1
H
O
H
C
R
C
C
CH3
step 2
H
H A
152
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
O
H
C
C
R
H H
Chapter 12: α-carbon part I
fig 24
In the retro-aldol cleavage reaction the β-hydroxy group is deprotonated (step 1 above),
to form a carbonyl, at the same time pushing off the enolate carbon, which is now a
leaving group rather than a nucleophile.
Is an enolate a good enough leaving group for this step to be chemically reasonable?
Sure it is: the same stabilizing factors that explain why it can form as an intermediate in
the forward direction (resonance delocalization of the negative charge to the oxygen,
interaction with a zinc cation) also explain why it is a relatively weak base, and therefore
a relatively good leaving group (remember, weak base = good leaving group!). All we
need to do to finish the reaction off is reprotonate the enolate (step 2) to yield the starting
aldehyde, and we are back where we started.
The key thing to keep in mind when looking at a retro-aldol mechanism is that, when the
carbon-carbon bond breaks, the electrons must have 'some place to go' where they will be
stabilized by resonance. Most often, the substrate for a retro-aldol reaction is a βhydroxy aldehyde, ketone, ester, or thioester.
If the leaving electrons cannot be stabilized, a retro-aldol cleavage step is highly unlikely.
carbonyl is three carbons away!
O
:B
C
H
H H O
?
R
C
C
R
C
C
H
H
H
O
H
X
H H
R
C
C
CH2
R
X
O
electrons cannot be
stabilized by resonance no retro-aldol!
fig 26a
The fructose 1,6-bisphosphate aldolase reaction we saw in the previous section is an
excellent example of an enzyme whose metabolic role is to catalyze both the forward and
reverse (retro) directions of an aldol reaction. The same enzyme participates both as an
aldolase in the sugar-building gluconeogenesis pathway, and as a retro-aldolase in the
sugar breaking glycolysis pathway. We have already seen it in action as an aldolase in the
gluconeogenesis pathway. Here it is in the glycolytic direction, catalyzing the retroaldol cleavage of fructose bisphosphate into DHAP and GAP:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
153
Chapter 12: α-carbon part I
The fructose 1,6-bisphosphate aldolase reaction (retro-aldol direction)
O
OH
O
OP
O
OP
+
OP
OH
OP
OH
OH
DHAP
fructose-1,6-bisphosphate
OH
GAP
Mechanism:
O
OP
:B
H
O
O
OH
O
OP
step 2
step 1
OP
OH
OP
OH
O
A
OH
OP
OH
H
fig 25
fig 26
Exercise 12.8: Predict the products of a retro-aldol reaction with the given substrate, and
draw the mechanism.
O HO CH3
CoAS
CO2
fig 27
12.3D: Aldol addition reactions with enzyme-linked enamine intermediates
Earlier we looked at the mechanism for the fructose 1,6-bisphosphate aldolase reaction in
bacteria. Interestingly, it appears that the enzyme catalyzing the exact same reaction in
plants and animals evolved differently: instead of going through a zinc-stabilized enolate
intermediate, in plants and animals the key intermediate is an enamine. The nucleophilic
substrate is first linked to the enzyme through the formation of an iminium with a lysine
residue in the enzyme's active site (refer to section 10.5 for the mechanism of iminium
formation). This effectively forms an 'electron sink', in which the positively-charged
iminium nitrogen plays the same role as the Zn+2 ion in the bacterial enzyme.
154
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
Mechanism for an aldol addition reaction with an enzyme-linked enamine intermediate
enz
H A
enz
NH3
enz
iminium
H2O
O
R
NH
R
iminium formation
B:
H
OH
OH
O
enamine
NH
step 1
H
H
R
R
OH
step 2
enz
enz
NH3
H2O
O
iminium
OH
R
R
NH
OH
iminium hydrolysis
R
OH
R
OH
fig 28
The α-proton, made more acidic by the electron-withdrawing effect of the imminium
nitrogen, is then abstracted by an active site base to form an enamine (step 1). In step 2 ,
the α-carbon attacks the carbonyl carbon of an aldehyde, and the new carbon-carbon
bond is formed. In order to release the product from the enzyme active site and free the
enzyme to catalyze another reaction, the iminium is hydrolyzed back to a ketone group
(see section 10.5 to review the imine/imminium hydrolysis mechanism).
There are many more examples of aldol/retroaldol reactions in which the key
intermediate is a lysine-linked imine. Many bacteria are able to incorporate
formaldehyde, a toxic compound, into carbohydrate metabolism by linking it to ribulose
monophosphate. The reaction (EC 4.1.2.43) proceeds through imine and enamine
intermediates.
OH
O
OH
O
PO
OH
OH
PO
+
H
H
HO H
HO H
ribulose-5-phosphate
O
formaldehyde
OH
hexulose-6-phosphate
fig 29
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
155
Chapter 12: α-carbon part I
Exercise 12.9: Propose a complete mechanism for the hexulose-6-phosphate aldolase
reaction shown above, including the imine formation and hydrolysis steps.
Here is an example of an enamine intermediate retro-aldol reaction from bacterial
carbohydrate metabolism (EC 4.1.2.14). Notice that the structures are drawn here in the
Fischer projection notation - it is important to practice working with this drawing
convention, as biologists and biochemists use it extensively to illustrate carbohydrate
chemistry. Proc. Natl. Acad. Sci. 2001, 98, 3679
H
OP
OP
OH
OH
OH
O
H
CH3
O
O
CO2
CO2
fig 30
Exercise 12.10: Draw the carbon-carbon bond breaking step in the reaction above. Use
the Fischer projection notation.
Section 4: α-carbon reactions in the synthesis lab - kinetic vs. thermodynamic
enolates
While aldol addition reactions are widespread in biochemical pathways as a way of
forming carbon-carbon bonds, synthetic organic chemists working the lab also make use
of aldol-like reactions for the same purpose. Consider this reaction:
Li
O
N
O
H
Br
H
O
+ LiBr
156
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
fig 30b
Here, cyclopentanone is deprotonated at an α-carbon by lithium diisopropylamide
(LDA), a very strong base commonly used in the synthesis lab. Addition to the reaction
mixture of an electrophile in the form of a primary alkyl bromide results in formation of a
new carbon-carbon bond. Notice that this is a kind of 'SN2 variation' on the aldol addition
reactions we saw above, because the enolate nucleophile is attacking in SN2 fashion
rather than in a carbonyl addition fashion.
What would happen, though, if we started with 2-ethylcyclopentantone? Because the
starting ketone is no longer symmetrical, we could hypothetically obtain two different
products:
O
1) LDA
2) bromoethane
O
-78 oC
kinetic product
25 oC
1) KH
2) bromoethane
O
thermodynamic product
fig 30c
It turns out that we can control which product we get by selecting the base used in the
reaction, and the reaction temperature. If we use LDA and immerse the reaction flask in
a dry ice-acetone bath (-78 oC), we get mainly 2,5-diethyl cyclopentanone. If we use
potassium hydride (KH) and run the reaction at room temperature, we get mainly 2,2diethylcyclopentanone.
LDA is a very hindered base: the basic nitrogen atom is surrounded by two bulky
isopropyl groups, and thus it is more difficult for it to come into contact with an αproton. The α-protons on the less substituted side of 2-ethylcyclopentanone are less
hindered and more accessible to the base. In addition, the cold reaction temperature
means that the deprotonation step is irreversible: the system does not have enough energy
to overcome the energy barrier for the reverse (reprotonation) reaction. The less
substituted enolate forms faster, and once it forms it goes on to attack the bromoethane
rather then reversing back to the ketone form. Because it is the rate of enolate formation
that determines the major product under these conditions, we say that this reaction is
under kinetic control, and the less substituted enolate intermediate is called the kinetic
enolate.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
157
Chapter 12: α-carbon part I
O
O
LDA
-78 oC
kinetic enolate
(forms faster)
25 oC
KH
O
thermodynamic enolate
(more stable)
fig 30d
If, on the other hand, we use KH as a base, hindrance is no longer an issue because the
base is a hydride ion. We run this reaction at room temperature, so the system has
enough energy to overcome the energy barrier for re-protonation, and enolate formation
is reversible. The enolate in most abundance at equilibrium is therefore not the one that
forms fastest, but the one that is more stable. The more substituted enolate is more stable
(recall that alkenes are more stable when they are more substituted - the same idea
applies here). The more substituted enolate leads to the 2,2-diethyl cyclopentanone
product. Because it is the stability of the enolate intermediate that determines the major
product under these conditions, we say that this reaction is under thermodynamic
control, and the more substituted enolate intermediate is the thermodynamic enolate.
158
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
Key learning objectives for this chapter
Before moving on to the next chapter, you should:
Understand what is meant by 'α and β positions' relative to a carbonyl group.
Understand how an enzyme can increase the acidity of an α-proton through the active site
microenvironment
Understand the 3D bonding arrangement of an enolate ion
Be able to recognize and draw reasonable mechanisms for the following reaction types:
tautomerizations: keto-enol, imine-enamine
racemization/epimerization
carbonyl isomerization (changing position of a carbonyl group)
alkene isomerization (changing position of an alkene relative to carbonyl)
aldol addition, retro-aldol cleavage (both enolate intermediate and enamine
intermediate mechanisms)
Be able to draw a mechanism for a laboratory alkylation reaction at the α-carbon of a
ketone or aldehyde. Understand the difference between kinetic and thermodynamic
control of this reaction type, and be able to predict the regiochemical outcome of the
reaction based on reaction conditions.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
159
Chapter 12: α-carbon part I
Problems
P12.1: The enzyme ribulose-5-phosphate isomerase (EC 5.3.1.6), which is active in both
the Calvin cycle and the pentose phosphate pathway, catalyzes an aldehyde-to-ketone
isomerization between two five-carbon sugars.
a) Draw a mechanism for this step.
OH
O
O
OH
OH
OH
OH
OH
OP
OP
ribulose 5-phosphate
ribose 5-phosphate
b) What 1H-NMR signal would most clearly differentiate the aldose from the ketose in
this reaction?
P12.2: Provide a likely mechanism for the reaction below, from tryptophan biosynthesis.
Hint: this is mechanistically very similar to a carbonyl isomerization reaction.
CO2 HO
N
H
CO2 HO
OP
OH
N
OH
H
OP
OH
O
P12.3:
a) Draw the product of an aldol addition reaction between pyruvate and glyoxylate (EC
4.1.3.16):
O
O
+
O
O
pyruvate
160
O
CH3
O
glyoxylate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
b) Draw the product of an aldol addition reaction between two molecules of pyruvate (EC
4.1.3.17):
c) The molecule below undergoes a retroaldol cleavage reaction in E. coli (J. Biol. Chem.
2012, 287, 36208). Draw the structure of the products.
OH
O
O2C
CO2
d) Propose a mechanism for this early reaction in the biosynthesis of isoprenoids. Hint:
this is an aldol reaction, followed by thioester hydrolysis.
acetyl CoA
H2O
O
O HO CH3 O
O
enz
O
S
SCoA
e) The carbon-carbon bond cleaving reaction below was reported to take place in many
species of bacteria. Predict the structure of product X, and draw a mechanism for the
reaction. Assume that an imine linkage to an enzymatic lysine residue does not play a
part in the mechanism. (J. Bacteriol. 2009, 191, 4158).
O
OH
N
HN
H2N
OH
N
N
H
X
+
HO
O
OH
P12.4: Below is a step in the biosynthesis of tryptophan. Draw a likely mechanism.
Hint: you will need to show an enamine to imine tautomerization step first, then the
carbon-carbon bond breaking step will become possible.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
161
Chapter 12: α-carbon part I
HO
OH
OH
PO
+
N
N
O
H
indole
H
OP
GAP
P12.5: The following step in the biosynthesis of lysine makes a connection between
aspartate semialdehyde and a common metabolic intermediate. Identify the intermediate,
and propose a mechanism for the reaction.
O
O
?
H
O
H3N
OH
O
O
O
H3N
O
O
aspartate semialdehyde
P12.6: Sugar-interconverting transaldol reactions play an important role in sugar
metabolism. In a transaldolase reaction, a ketose (e.g. fructose-6-phosphate) first breaks
apart in a retro-aldol step to release an aldose (e.g. glyceraldehyde-3-phosphate) from the
active site. Then, in a forward aldol step, a second aldose (e.g. erythrose-4-phosphate)
enters the active site and connects to what remains from the original ketose (the red part
in the figure below) to form a new ketose (e.g. sedoheptulose-7-phosphate).
Transaldolase enzymes generally have a lysine in the active site that is covalently bound
to the substrate throughout the reaction cycle.
Draw curved-arrow diagrams showing a) the carbon-carbon bond breaking step of the
reaction cycle, and b) the carbon-carbon bond forming step.
OH
OH
O
O
HO
OH
OH
OP
fructose-6phosphate
162
O
+
OH
OH
OP
erythrose-4phosphate
HO
O
OH
+
OP
glyceraldehyde3-phosphate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OH
OH
OH
OP
sedoheptulose-7phosphate
Chapter 12: α-carbon part I
12.7: Scientists are investigating the enzymatic reaction below, which is part of the
biosynthesis of the outer membrane of gram-negative bacteria, as a potential target for
new antibiotic drugs. Draw a likely mechanism for the reaction. (J. Biol. Chem. 2008,
283, 2835).
OH
OH
OH
O
PO
OH
O
PO
OH
OH
OH
OH
OH
OH
12.8: The reaction below, catalyzed by the enzyme malate synthase, is part of the
glyoxylate cycle of plants and some bacteria. It is the glyoxylate cycle that allows these
organisms to convert acetyl CoA, derived from the metabolism of oils, into glucose.
O
CoASH
H2O
acetyl-CoA
O
OH
O
H
O
O
O
O
glyoxylate
malate
a) Propose a mechanism.
b) Predict the signals you would expect to see in a 1H-NMR spectrum of malate.
P12.9 The reaction below, from the biosynthetic pathway for the amino acid tryptophan,
is dependent upon a coenzyme that we learned about in an earlier chapter. Based on the
reaction, identify this coenzyme and propose a mechanism.
OH
O
H3C
CO2
CO2
(coenzyme)
N
N
H
H
P12.10: In the biosynthesis of leucine, acetyl CoA condenses with another metabolic
intermediate ‘X” to form 1-isopropylmalate (EC 2.3.3.13). Give the structure for
substrate X, and provide a mechanism for the reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
163
Chapter 12: α-carbon part I
H2O
acetyl CoA
HSCoA
CO2
X
HO CO2
P12.11:
a) Mycobacterium tuberculosis, the microbe that causes tuberculosis, derives energy
from the metabolism of cholesterol from infected patients. The compound below is
predicted to be an intermediate in that metabolic pathway, and to undergo a retro-aldol
cleavage reaction. Predict the retro-aldol products and show the mechanism involved.
(Crit. Rev. Biochem. Mol. Biol. 2014, 49, 269, fig 5).
O
OH
SCoA
retro-aldol products
R
R
H
b) Polyketides are a structurally diverse class of biomolecules produced by almost all
living things. Many drugs are derived from polyketide precursors. The cancer drug
doxorubicin (trade name Adriamycin) is derived from a bacterial polyketide called
rhodomycinone. Aklaviketone, an intermediate in the biosynthesis of rhodomycinone, is
derived in a single enzymatic step from akalonic methyl ester, in a reaction in which the
carbon-carbon bond indicated by an arrow is formed. Predict the structure of akalonic
methyl ester.
O
O
OCH3
OH
akalonic methyl ester
OH
O
OH
aklaviketone
P12.12: The unusual isomerization reaction shown below has been reported recently to
occur in some bacteria. Propose a mechanism that begins with formation of an enolate
intermediate. (J. Biol. Chem. 2012, 287, 37986).
164
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 12: α-carbon part I
O
SCoA
SCoA
O
O
O
P12.13: The reactions in parts a) and b) below both proceed through lysine-linked
enamine intermediates. Show the carbon-carbon bond forming step for each reaction.
Hint: you will want to consider the straight-chain (ie. aldose/ketose) form of the sugars in
both cases.
a) (J. Bacteriol. 2004, 186, 4185)
OH
O
H3C
+
H
PO
O
OP
acetaldehyde
OH
O
GAP
OH
deoxyribose 5-phosphate
b) (J. Biol. Chem. 2011, 286, 14057)
HO
CH3
HO
HO
HO
HN
O
O
OH
O
+
H3C
CO2
OH
OH
HO
H
N
H3C
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
O
CO2
HO
165
Interchapter: Pathway prediction
Interchapter
(between chapter 12 and 13)
Predicting metabolic pathways
Imagine that you are a biological chemist doing research on bacterial metabolism. You
and your colleagues isolate an interesting biomolecule from a bacterial culture, then use
mass spectrometry, NMR, and other analytical techniques to determine its structure.
Using your 'toolbox' of known organic reaction types - nucleophilic substitution,
phosphorylation, aldol additions, and so forth - can you figure out a chemically
reasonable pathway by which your compound might be enzymatically synthesized from
simple metabolic precursors? In other words, can you fill in the missing biochemical
steps (or at least some of them) to come up with a potential new metabolic pathway,
which can then be used a hypothesis for future experimental work to try to find and
study the actual enzymes involved?
An actual example approximating this scenario is shown below. A complete biosynthetic
pathway for isopentenyl diphosphate (IPP), the building block molecule for all isoprenoid
compounds (section 1.3A), has been known since the 1960's. This pathway, which
begins with acetyl-CoA, was shown to be active in yeast, plants, and many other species
including humans. However, researchers in the late 1980s uncovered evidence indicating
that the known pathway is not present in bacteria, although they clearly use IPP as a
building block molecule just as other forms of life do.
166
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Interchapter: Pathway prediction
O
H3C
C
SCoA
3X acetyl-CoA
known pathway
(in humans, yeast, etc.)
PPO
common pathway
all isoprenoid compounds
isopentenyl diphosphate
(IPP)
unknown pathway
(in bacteria)
OH
O
+
OP
O
glyceraldehyde phosphate
(GAP)
H3C
C
C
O
O
pyruvate
fig 1a
Over the next several years, the researchers conducted a number of experiments in which
bacteria were grown on a medium containing glucose 'labeled' with the 13C isotope.
With the results from these experiments, combined with their knowledge of common
biological organic reaction types, the researchers were able to correctly predict that the
bacterial pathway starts with two precursor molecules (pyruvate and glyceraldehyde
phosphate instead of acetyl CoA) and they also correctly predicted the first two
enzymatic steps of the newly discovered bacterial pathway. This accomplishment
eventually led to elucidation of every step in the pathway, and isolation of the enzymes
catalyzing them. (Biochem J. 1993, 295, 517; J. Am. Chem. Soc. 1996, 118, 2564; Lipids
2008, 43, 1095)
Why weren't they able to predict the whole pathway? It turns out that several of the later
steps were somewhat unusual, unfamiliar reaction types - but discovery of these reactions
hinged upon the correct prediction of the more familiar first two steps.
Multi-step transformation problems of this type offer an unparalleled opportunity to use
our knowledge of biological organic chemistry combined with creative reasoning to solve
challenging, relevant scientific puzzles. At this point in your organic chemistry career,
you have not yet accumulated quite enough tools in your reaction toolbox to tackle most
real-life biochemical pathway problems such as the one addressed above - but by the time
we finish with oxidation and reduction chemistry in chapter 15, you will be able to
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
167
Interchapter: Pathway prediction
recognize most of the reaction types that you will encounter in real metabolism, and will
be challenged to predict some real pathways in the end-of-chapter problems.
You do, however, have right now enough of a bioorganic repertoire to begin to learn how
multi-step pathway problems can be approached, using for practice some generalized,
hypothetical examples in which the reaction types involved are limited to those with
which you are already familiar.
Imagine that you want to figure out how an old-fashioned mechanical clock is put
together. One way to do this is to start with a working clock, and take it apart piece-bypiece. Alternatively, one could start with all of the disassembled pieces, plus a lot of
other small parts from different clocks, and try to figure out how to put together the
specific clock you are interested in. Which approach is easier? The answer is intuitively
obvious - it's usually easier to take things apart than to put them back together.
The same holds true for molecules. If we want to figure out the biosynthetic pathway by
which a large, complex biomolecule might be made in a cell, it makes sense to start with
the finished product and then mentally work backwards, taking it apart step-by-step using
known, familiar reactions, until we get to simpler precursor molecules. Starting with a
large collection of potential precursor molecules and trying to put the right ones together
to make the target product would be a formidable task.
Retrosynthetic analysis - the concept of mentally dismantling a molecule step by step all
the way back to smaller, simpler precursors using known reactions - is a powerful and
widely-used intellectual tool first developed by synthetic organic chemists. The approach
has also been adapted for use by biological chemists in efforts to predict pathways by
which known biomolecules could be synthesized (or degraded) in living things.
In retrosynthesis, we think about a series of reactions in reverse. A backwards (retro)
chemical step is symbolized by a 'thick' arrow, commonly referred to as a retrosynthetic
arrow, and visually conveys the phrase 'can be formed from'.
retrosynthetic (backwards) arrow
+
'can be formed from'
target molecule
two simpler precuror molecules
fig 2a
Consider a simple, hypothetical example: starting with the target molecule below, can we
come up with a chemically reasonable pathway starting from the precursors indicated?
168
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Interchapter: Pathway prediction
O
O
O
CH3
O
+
R1
CH3
H
R2
R1
R2
disconnection
fig 2b
A first step is to identify the relevant disconnection: a key bond (usually a carbon-carbon
bond) that must be formed to make the target product from smaller precursors. We search
our mental 'toolbox' of common biochemical reaction types, and remember that the only
way we know of (so far!) to make a new carbon-carbon bond is through an aldol addition
reaction, which takes place at an α-carbon. Therefore, we can make a likely
disconnection next to the α-carbon in the target molecule.
Next, we need to recognize that the aldol addition reaction results in a β-hydroxy ketone.
But our target molecule is a β-methoxy ketone! Working backwards, we realize that the
β-methoxy group could be formed from a β-hydroxy group by a SAM methylation
reaction (section 8.8A). This is our first retrosynthetic (backwards) step.
O
O
R1
CH3
O
'can be
formed from'
R2
OH
R1
(+ SAM)
R2
fig 2c
The second retro step (aldol) accounts for the disconnection we recognized earlier, and
leads to the two precursor molecules.
O
O
O
OH
+
R1
R2
'can be
formed from'
R1
CH3
H
R2
fig 2d
Now, consider the more involved (but still hypothetical) biochemical transformation
below:
OH
O
OH
R1
+
HO
R2
O
A
R1 OH
OH
b
c
HO
B
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
a
O
e
d
R2
OCH3
f
169
Interchapter: Pathway prediction
fig 3
Often the best thing to do first in this type of problem is to count the carbons in the
precursor compounds and product - this allows us to recognize when extra carbons on
either side must at some point be accounted for in our solution. In this case, one carbon
(labeled 'f'') has been gained in the form of a methyl ether in the product. This is easy to
account for: we know that the coenzyme S-adenosyl methionine (SAM - section 8.8A)
often serves as the methyl group donor in enzymatic O- or N-methylation reactions. So,
we can propose our first backwards (retro) step: the product as shown could be derived
from SAM-dependent methylation of an alcohol group on a proposed intermediate I.
Retrosynthetic step 1:
R1 OH
HO
b
c
HO
a
R1 OH
HO
O
d
a
b
c
e
R2
retrosynthesis direction
d
HO
OCH3
I
product
R1 OH
b
c
HO
a
d
R2
OH
f
HO
O
e
SAM
SAH
R1 OH
HO
b
c
O
e
R2
forward synthesis direction
HO
a
O
e
d
R2
OH
OCH3
I
product
f
fig 4
How do we know that the methylation step occurs last? We don't - remember, we are
proposing a potential pathway, so the best we can do is propose steps that make chemical
sense, and which hopefully can be confirmed or invalidated later through actual
experimentation. For now, we'll stick with our initial choice to make the methylation
step the last one.
Now that we have accounted for the extra carbon, a key thing to recognize regarding the
transformation in question is that two linear molecules are combining to form a cyclic
product. Thus, two connections need to be made between reactants A and B, one to join
the two, the other to close the circle. Our primary job in the retro direction, then, is to
establish in the product the two points of disconnection: in other words, to find the two
bonds in the product that need to be taken apart in our retrosynthetic analysis. Look
closely at the product: what functional groups do you see? Hopefully, you can identify
two alcohol groups, a methyl ether, and (critically) a cyclic hemiketal. We've already
accounted for the methyl ether. Identifying the cyclic hemiketal is important because it
170
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Interchapter: Pathway prediction
allows us to make our next 'disconnection': we know how a hemiketal forms from a
ketone and an alcohol (section 10.2), so we can mentally work backwards and predict the
open-chain intermediate II that could cyclize to form our product.
R1 OH
HO
b
c
HO
a
d
R1 O
HO
O
e
retrosynthesis direction
R2
HO
b
c
HO
a
d
OH
e
R2
OH
OH
I
II
R1 OH
R1 O
HO
b
c
a
d
HO
b
c
OH
e
forward synthesis direction
R2
HO
a
d
hemiketal
O
e
R2
OH
OH
II
I
fig 5
Now, starting with the R1 group and working along the carbon chain, we can account for
carbons a-e on the two precursors.
OH
O
R1
OH
a
+
c
R1 O
OH
d
e
b
HO b
R2
O
A
B
starting precursors
HO
c
a
OH
d
e
R2
OH
intermediate II
fig 5a
Thus, the next disconnection is between carbons b and c. Here's where our mastery of
biological organic reactivity really comes into play: the OH at carbon c of intermediate II
is in the β position relative to carbonyl carbon a. Aldol addition reactions (section 12.3)
result in β-hydroxy ketones or aldehydes. Therefore, we can work backward one more
step and predict that our intermediate II was formed from an aldol addition reaction
between intermediate III (as the nucleophile) and precursor molecule A (as the
electrophile).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
171
Interchapter: Pathway prediction
R1 O
R1 O
HO
b
a
b
c
d
HO
HO
OH
+
e
R2
OH
retrosynthesis direction
O
c
d
R1 O
HO
H H
+
c
O
d
III
R2
R1 O
A
a
OH
e
OH
II
HO b
A
a
forward synthesis direction
b
c
HO
a
d
OH
e
R2
OH
OH
e
R2
III
II
OH
fig 6
We are most of the way home - we have successfully accounted for given precursor A.
Intermediate III, however, is not precursor B. What is different? Both III and B have a
carbonyl and two alcohol groups, but the positioning is different: III is an aldehyde, while
B is a ketone. Think back to earlier in this chapter: intermediate III could form from
isomerization of the carbonyl group in compound B (section 12.2A). We have now
accounted for our second precursor - we are done!
172
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Interchapter: Pathway prediction
OH
OH
c
d
O
c
e
R2
OH
d
HO
retrosynthesis direction
B
OH
OH
c
R2
O
III
HO
e
d
c
e
R2
d
O
forward synthesis direction
e
O
OH
B
III
R2
fig 7
In the forward direction, a complete pathway diagram can be written as follows:
R1
HO
O
A
OH
OH
R2
HO
O
carbonyl
isomerization
R2
O
R1 O
HO
aldol addition
OH
R2
HO
OH
OH
II
III
B
hemiketal
formation
R1 OH
R1 OH
HO
SAH
SAM
HO
O
O
R2
HO
O-methylation
OCH3
product
R2
HO
OH
I
fig 8
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
173
Interchapter: Pathway prediction
A full 'retrosynthesis' diagram for this problem looks like this:
R1 OH
HO
R1 OH
HO
O
R2
HO
R1 O
HO
O
R2
HO
OCH3
OH
product
I
OH
R2
HO
OH
II
R1
HO
+
OH
R2
HO
O
A
OH
R2
O
O
OH
B
IV
fig 9
In the multi-step pathway prediction problems that you will be asked to solve below and
in the remainder of this book, you will be instructed to present your solution in the form
of a proposed 'forward' pathway diagram, showing the participation of all coenzymes and
other species such as water. At first, we'll start with relatively simple, hypothetical
biochemical transformations. As you learn more reaction types in chapter 13-17, the
range and complexity of problems that you will be able to solve will expand
correspondingly, and you will eventually be able to tackle real-life pathways.
174
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Interchapter: Pathway prediction
Problems
For each transformation below, draw a pathway diagram illustrating a potential
biosynthetic pathway. Indicate other molecules participating in the reaction but not
shown below (eg. coenzymes, water, etc.). Each step should be recognizable as a reaction
type that we have covered through the end of chapter 12. (Note - you are being asked to
draw your pathways in the 'forward' direction, but you should attack each problem using
a retrosynthetic analysis strategy).
1:
O
O
HO
HO
+
O
O
O
O
OH
HO
O
2:
allylic alcohol
+
aldehyde
+
ammonia
R2
N
R1
3:
OH
OH
OH
O
+
R
R
one of the 20 common
amino acids
N
O2C
OH
NH3
4:
O
R 1
H2N
OH
2
3
7
4
7
OH
N
O
6
5
3
OH
R
1
4
OH
6
5
2
OH
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OH
OH
175
Interchapter: Pathway prediction
5:
O
O
OH
O
OH
+
+ RNH
3
OH
CO2
NH
O
R
6:
O
HO
HO
176
OCH3
NH3
OH
O
HO
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OH
Chapter 13: α-carbon part II
Chapter 13
Reactions at the α-carbon, part II
Introduction
We begin this chapter with the story of two men, and two chemical reactions.
The two men couldn't be more different. One was an acclaimed scientist who lived and
continued to work productively into his eighties. The other was struck down as a young
boy by what was assumed at the time to be a fatal disease. With the heroic support of his
parents and caregivers, though, he lived to his thirtieth birthday and provided the
inspiration for development of a medical treatment that could potentially save thousands
of lives.
The two chemical reactions in this story are closely related, and both involve the
metabolism of fats in the human body. One serves to build up fatty acid chains by
repeatedly linking together two-carbon units, while the other does the reverse,
progressively breaking off two-carbon pieces from a long chain fatty acid molecule. The
life and work of the two men are inextricably linked to the two reactions, and while we
will be learning all about the reactions in the main part of this chapter, we'll begin with
the stories of the two men.
On a Saturday in January of 2007, Dr. Hugo Moser passed away in the Johns Hopkins
Hospital in Baltimore, succumbing to pancreatic cancer. He was 82 years old. A
neurologist who had taught and researched for much of his career at Johns Hopkins, he
was well known for his workaholic nature: he had signed off on his last grant application
while on the way to the hospital for major surgery just a few months previously. Two
days after his death, his wife and colleague Ann Moser was back in their lab, because,
she said, “He gave us all a mandate to continue with the work”. Dr. Moser was a highly
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
177
Chapter 13: α-carbon part II
esteemed scientist who had devoted his life to understanding and eventually curing a
class of devastating neurodegenerative diseases, most notably adrenoleukodystrophy, or
ALD. In his work he was careful, rational, painstaking, and relentless – a classic scientist.
But in the minds of many movie fans, he became a Hollywood villain.
Only 17 months after the death of Dr. Moser, newspapers around the world published
moving obituaries marking the passing, at age 30, of Lorenzo Odone. In one, written by
his older sister and published in the British newspaper The Guardian, Lorenzo as a young
boy is described as “lively and charming . . .he displayed a precocious gift for languages
as he mastered English, Italian and French. He was funny, articulate and favored opera
over nursery rhymes.” But for more than 20 years leading up to his death, he had been
confined to a wheelchair, blind, paralyzed, and unable to communicate except by
blinking his eyes. Because he was unable to swallow, he needed an attendant to be with
him around the clock to suction saliva from his mouth so he wouldn't choke.
When he was he six years old, Lorenzo started to show changes in behavior: a shortening
attention span, moodiness. More disturbing to his parents, Augusto and Michaela Odone,
was their suspicion that he was having trouble hearing. They took him in to be examined,
and although his hearing was fine, the doctors noticed other behavioral symptoms that
concerned them, and so ordered more neurological tests. The results were a kick to the
stomach: Lorenzo had a fatal neurodegenerative disease called adrenoleukodystrophy.
There was no cure; his nervous system would continue to degenerate, and he would
probably be dead within two years.
What happened next became such a compelling story that it was eventually retold by
director George Miller in the 1992 movie Lorenzo's Oil, starring Nick Nolte and Susan
Sarandon as Augusto and Michaela Odone and Peter Ustinov as a character based on Dr.
Hugo Moser. The Odones were unwilling to accept the death sentence for their son and,
despite having no scientific or medical training, set about to learn everything they could
about ALD.
They found out that the cause of ALD is a mutation in a gene that plays an important role
in the process by which saturated fatty acids of 26 or more carbons are broken down in
the body. When these 'very long chain fatty acids' (VLCFAs) accumulate to excessive
levels, they begin to disrupt the structure of the myelin sheath, a protective fatty coating
around nerve axons, leading eventually to degradation of the nervous system.
Researchers had found that restricting dietary intake of VLCFAs did not help –
apparently much of the damage is done by the fats that are naturally synthesized by the
body from shorter precursors. The Odones realized that the key to preventing destruction
of the myelin sheath might be to somehow disrupt the synthesis of VLCFAs in Lorenzo's
cells. The breakthrough came when they came across studies showing that the carbon
chain-elongating enzyme responsible for producing VLCFAs is inhibited by oleic and
erucic acids, which are monounsaturated fatty acids of 18 and 22 carbons, respectively
and are found in vegetable oils.
178
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
O
26
OH
1
16
a 26-carbon VLCFA
O
9
1
OH
oleic acid
18
O
13
1
OH
erucic acid
22
fig 38a
Administration of a mixture of these two oils, which eventually came to be known as
'Lorenzo's Oil' , was shown to lead to a marked decrease in levels of VLCFAs in ALD
patients.
This was, however, a therapy rather than a miracle cure – and tragically for the Odones
and the families of other children afflicted with ALD, the oil did not do anything to
reverse the neurological damage that had already taken place in Lorenzo's brain.
Although he was profoundly disabled, with round-the-clock care and a daily dosage of
the oil Lorenzo was able to live until a day after his 30th birthday, 22 years longer than his
doctors had predicted.
The story does not end there. Although the discovery of the treatment that bears his name
came too late for Lorenzo Odone, might daily consumption of the oil by young children
who are at a high genetic risk for ALD possibly prevent onset of the disease in the first
place, allowing them to live normal lives? This proposal was not without a lot of
controversy. Many ALD experts were very skeptical of the Lorenzo's oil treatment as
there was no rigorous scientific evidence for its therapeutic effectiveness, and indeed
erucic oil was thought to be potentially toxic in the quantities ingested by Lorenzo. Most
doctors declined to prescribe the oil for their ALD patients until more studies could be
carried out. The Hollywood version of Lorenzo's story cast the medical and scientific
establishment, and Dr. Hugo Moser in particular, in a strikingly negative light – they
were portrayed as rigid, callous technocrats who cared more about money and academic
prestige than the lives of real people. Dr. Moser was not mentioned by name in the
movie, but the character played by Peter Ustinov was based closely on him: as his
obituary in the Washington Post recounts, Dr. Moser once told an interviewer "The good
guys were given real names. The bad guys were given pseudonyms."
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
179
Chapter 13: α-carbon part II
What Hugo Moser in fact did was what a good scientist should always do: he kept an
open mind, set up and performed careful, rigorous experiments, and looked at what the
evidence told him. In a 2005 paper, Moser was finally able to confidently report his
results: when young children at risk of developing ALD were given a daily dose of
Lorenzo's oil, they had significantly better chance of avoiding the disease later on.
When he died, Dr. Moser was tantalizingly close to demonstrating conclusively that a
simple and rapid blood test that he and his team had developed could reliably identify
newborns at high risk of developing ALD – but it was not until after his death that his
colleagues, including his wife, Ann Moser, were able to publish results showing that the
test worked. The hope is that many lives might be saved by routinely screening newborns
for ALD and responding with appropriate preventive treatments - possibly including
Lorenzo's oil.
The biochemical reactions at the heart of the Lorenzo's oil story – the carbon-carbon
bond forming and bond breaking steps in the synthesis and degradation of fatty acids both involve chemistry at the α-carbon and proceed through enolate intermediates, much
like the aldol and isomerization reactions we studied in chapter 11. They are known as
'Claisen condensation' and 'retro-Claisen cleavage' reactions, respectively, and represent
another basic mechanistic pattern - in addition to the aldol reaction - that is ubiquitous in
metabolism as a means of forming or breaking carbon-carbon bonds.
To begin this chapter, we will first learn about 'carboxylation' and 'decarboxylation'
reactions, in which organic molecules gain or lose a bond to carbon dioxide, respectively,
in a mechanism that is really just an extension of the aldol/retro-aldol reactions we
learned about in the previous chapter. As part of this discussion, we will work through
the mechanism of the carbon-fixing enzyme in plants commonly known as 'Rubisco',
which is thought to be the most abundant enzyme on the planet. Then, we will move to
the Claisen reactions that are so central to lipid metabolism and the story of Lorenzo
Odone. Finally, we will study 'conjugate additions' and 'β-eliminations', common reaction
patterns that involve double bonds in the α−β position relative to a carbonyl group, and
which, again, proceed via enolate intermediates.
Section 13.1: Decarboxylation
Many carbon-carbon bond-forming and bond-breaking processes in biological chemistry
involve the gain or loss, by an organic molecule, of a single carbon atom in the form of
CO2. You undoubtedly have seen this chemical equation before in an introductory
biology or chemistry class:
6CO2(g) + 6H2O(l) + energy → C6H12O6(aq) + 6O2(g)
This of course represents the photosynthetic process, by which plants (and some bacteria)
harness energy from the sun to build glucose from individual carbon dioxide molecules.
The key chemical step in plants in which a carbon dioxide molecule is 'fixed' (linked to a
180
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
larger organic molecule) is a carboxylation reaction, and is catalyzed by the enzyme
ribulose 1,5-bisphosphate carboxylase, commonly known as Rubisco.
The reverse chemical equation is also probably familiar to you:
C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) + energy
This equation expresses what happens in the process known as respiration: the oxidative
breakdown of glucose to form carbon dioxide, water, and energy (in a non-biological
setting, it is also the equation for the uncatalyzed combustion of glucose). In respiration,
each of the carbon atoms of glucose is eventually converted to a CO2 molecule. The
process by which a carbon atom - in the form of carbon dioxide - breaks off from a larger
organic molecule is called decarboxylation.
We will look now at the biochemical mechanism of decarboxylation reactions. Later in
the chapter, we will look at the carboxylation reaction catalyzed by the Rubisco enzyme.
Decarboxylation steps occur at many points throughout central metabolism. Most often,
the substrate for a decarboxylation step is a β-carboxy ketone or aldehyde.
Decarboxylation of a β-carboxy ketone or aldehyde:
O
R
C
H
C
α
C
β
O
CO2
O
R
O
H
C
CH3
Mechanism:
carboxylate is β to
carbonyl
R
O
O
O
C
C
C
C β O
α
H
H
R
H A
O
C
H
H
R
+
C
CH3
O
C
O
fig 40
Just as in a retro-aldol reaction, a carbon-carbon bond is broken, and the electrons from
the broken bond must be stabilized for the step to take place. Quite often, the electrons
are stabilized by the formation of an enolate, as is the case in the general mechanism
pictured above.
The electrons from the breaking carbon-carbon bond can also be stabilized by a
conjugated imine group and/or by a more extensively conjugated carbonyl.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
181
Chapter 13: α-carbon part II
A
R
H
N
O
R
O
O
C
R
O
O
fig 42
The key in understanding decarboxylation reactions is to first mentally 'push' the
electrons away from the carboxylate group, then ask yourself: where do these electrons
go? If the electrons cannot 'land' in a position where they are stabilized, usually by
resonance with an oxygen or nitrogen, then a decarboxylation is very unlikely.
The compound below is not likely to undergo decarboxylation:
no β-carbonyl
H OH ? O
C
C
R
C
O
H2
O
H OH
X
R
C
+
C
O
CH2
electrons cannot be
stabilized by resonance no decarboxylation!
fig 41
Be especially careful, when drawing decarboxylation mechanisms, to resist the
temptation to treat the CO2 molecule as the leaving group in a mechanistic sense:
Not a decarboxylation mechanism!
R
O
O
C
C
C
H H
O
O
X
R
C
O
C
H
+
C
O
H
???
highly unstable products!
fig 43
The above is not what a decarboxylation looks like! (Many a point has been deducted
from an organic chemistry exam for precisely this mistake!) Remember that in a
decarboxylation step, it is the rest of the molecule that is, in fact, the leaving group,
'pushed off' by the electrons on the carboxylate.
182
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
Decarboxylation reactions are generally thermodynamically favorable due to the entropic
factor: one molecule is converted into two, one of which is a gas - this represents an
increase in disorder (entropy). Enzymatic decarboxylation steps in metabolic pathways
are also generally irreversible.
Below are two decarboxylation steps (EC 1.1.1.42; EC 1.1.1.43) in central catabolic
metabolism (specifically the citric acid cycle and pentose phosphate pathway catabolism,
respectively). Each step representing a point at which a carbon atom derived from the
food we eat is released as carbon dioxide:
H+
O
O2C
CO2
O
O2C
CO2
CO2
CO2
CO2
OH
O
H+
CO2
OH
O
OH
OH
OH
OH
OP
OP
fig 44
Exercise 13.1:
a) Draw mechanistic arrows showing the carbon-carbon bond breaking step in each of the
reactions shown above.
b) For the top reaction, provide a sound argument for why the other two carboxylate
groups are unlikely to be lost.
The reaction catalyzed by acetoacetate decarboxylase (EC 4.1.1.4) relies on an imminium
(protonated imine) link that forms temporarily between the substrate and a lysine residue
in the active site, in a strategy that is similar to that of the enamine-intermediate aldolase
reactions we saw in chapter 12. (Recall from section 7.5 that the pKa of an imminium
cation is approximately 7, so it is generally accurate to draw either the protonated
imminium or the neutral imine in a biological organic mechanism).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
183
Chapter 13: α-carbon part II
enz
enz
H3N
H3N
enz
H2O
O
CO2
enz
CO2
HN
H2O
O
HN
CO2
acetone
acetoacetate
fig 45
Exercise 13.2: Draw a mechanism for the carbon-carbon bond breaking step in the
acetoacetate decarboxylase reaction.
Exercise 13.3: Which of the following compounds could be expected to potentially
undergo decarboxylation? Draw the mechanistic arrows for the decarboxylation step of
each one you choose, showing how the electrons from the breaking carbon-carbon bond
can be stabilized by resonance.
O
a)
O2C
CO2
O
O2C
SCoA
O
HO CH3 O
c)
O2C
b)
CO2
d)
CO2
SCoA
N
OP
OH
N
CH3
H
fig45a
Section 13.2: An overview of fatty acid metabolism
In the introduction to this chapter, we learned about a patient suffering from a rare
disease affecting fatty acid metabolism. The reaction mechanisms that we are about to
learn about in the next two sections are central to the process by which fatty acids are
assembled (synthesis) and taken apart (degradation), so it is worth our time to go through
a brief overview before diving into the chemical details.
Fatty acid metabolism is a two-carbon process: in the synthetic directions, two carbons
are added at a time to a growing fatty acid chain, and in the degradative direction, two
184
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
carbons are removed at a time. In each case, there is a four-step reaction cycle that gets
repeated over and over. We will learn in this chapter about steps I and III in the synthesis
direction, and steps II and IV in the degradative direction. The remaining reactions, and
the roles played by the coenzymes involved, are the main topic of chapter 15.
Fatty acid synthesis:
'ACP' stands for 'acyl carrier protein', which is a protein that links to growing fatty acid
chains through a thioester group (see section 11.5A)
Step I: Condensation (covered in section 13.3A)
O
R
CO2
HSACP
O
SACP
+
O2C
2
growing fatty acid
(linked to acyl carrier protein)
1
O
R
SACP
O
SACP
two carbons added
malonyl-ACP
Step II: Ketone hydrogenation (covered in section 15.3)
NADPH
O
NADP+
OH
O
R
O
R
SACP
SACP
Step III: Elimination (covered in section 13.4)
OH
H2O
O
R
O
R
SACP
SACP
Step IV: Alkene hydrogenation (covered in section 15.4)
NADPH
O
R
SACP
NADP+
O
R
SACP
. . . back to step I, add another malonyl-ACP, repeat.
fig 31a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
185
Chapter 13: α-carbon part II
Fatty acid degradation:
Step I: Alkane oxidation (covered in section 15.4)
R
FADH2
FAD
O
O
R
SCoA
SCoA
fatty acid thioester
Step II: Addition of water (covered in section 13.4)
H2O
O
R
OH
SCoA
O
R
SCoA
Step III: Oxidation of alcohol (covered in section 15.3)
OH
NAD+
O
R
NADH
O
O
R
SCoA
SCoA
Step IV: Cleavage (covered in section 13.3C)
O
O
R
HSCoA
O
SCoA
O
R
SCoA
+
H3C
SCoA
acetyl-CoA
fatty acid thioester is shorter
by two carbons
. . . back to step I
fig 31b
When looking at these two pathways, it is important to recognize that they are not the
reverse of each other. Different coenzymes are in play, different thioesters are involved
(coenzyme A in the degradative direction, acyl carrier protein in the synthetic direction),
and even the stereochemistry is different (compare the alcohols in steps II/III of both
pathways). As you will learn in more detail in a biochemistry course, metabolic pathways
that work in opposite directions are generally not the exact reverse of each other. In
some, like fatty acid biosynthesis, all of the steps are catalyzed by different enzymes in
the synthetic and degradative directions. Other 'opposite direction' pathways, such as
glycolysis/gluconeogenesis, contain mostly reversible reactions (each catalyzed by one
enzyme working in both directions), and a few irreversible 'check points' where the
reaction steps are different in the two directions. As you will learn when you study
186
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
metabolism in biochemistry course, this has important implications in how two 'opposite
direction' metabolic pathways can be regulated independently of one another.
Recall that in Chapter 12 we emphasized the importance of two reaction types - the aldol
addition and the Claisen condensation - in their role in forming (and breaking) most of
the carbon-carbon bonds in a living cell. We have already learned about the aldol
addition, and its reverse, the retro-aldol cleavage. Now, we will study the Claisen
condensation reaction, and its reverse, the retro-Claisen cleavage. Step I in fatty acid
synthesis is a Claisen condensation, and step IV in fatty acid degradation is a retroClaisen cleavage.
In section 13.4, we will look more closely at the reactions taking place in step III of fatty
acid synthesis (an elimination) and step II of fatty acid degradation (a conjugate addition)
Section 13.3: Claisen condensation
13.3A: Claisen condensation - an overview
Recall the general mechanism for a nucleophilic acyl substitution mechanism, which we
studied in chapter 10:
O
B:
H Nu
R
C
O
O
X
R C
X
Nu
R
C
Nu
+
H X
H A
fig 31
The major points to recall are that a nucleophile attacks a carboxylic acid derivative,
leading to a tetrahedral intermediate, which then collapses to expel the leaving group (X).
The whole process results in the formation of a different carboxylic acid derivative.
A typical nucleophilic acyl substitution reaction might have an alcohol nucleophile
attacking a thioester, driving off a thiol and producing an ester.
O
O
B:
C
H OR
R
alcohol
thioester
SR
O
R C
SR
RO
H A
R
C
OR
ester
+
RSH
thiol
tetrahedral
intermediate
fig 32
If, however, the attacking nucleophile in an acyl substitution reaction is the α-carbon of
an enolate, a new carbon-carbon bond is formed. This type of reaction is called a
Claisen condensation, after the German chemist Ludwig Claisen (1851-1930).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
187
Chapter 13: α-carbon part II
A Claisen condensation reaction
O
RS
C
O
CH3
+
RS
C
O
R
RS
C
O
C
C
R
+
RSH
H H
thioester 1
thioester 2
thiol
β-ketothioester
Mechanism:
O
RS
C
C
H
H H
B:
step 1
A
O
RS
C
H
O
C
H
H
RS
C
new carbon-carbon bond
RS
R
step 2
O
RS
O
C R
C H
H
O
step 3
RS
C
O
C
C
R
+
RSH
H H
fig 33
In step 1, the α−carbon of one thioester substrate is deprotonated to form an enolate,
which then goes on to attack the second thioester substrate (step 2). Then the resulting
tetrahedral intermediate collapses (step 3), expelling the thiol leaving group and leaving
us with a β-keto thioester product (a thioester with a ketone group two carbons away).
To reiterate: A Claisen condensation reaction is simply a nucleophilic acyl substitution
(Chapter 11) reaction with an enolate carbon nucleophile.
13.3B: Biochemical Claisen condensation examples
A Claisen condensation between two acetyl CoA molecules (EC 2.3.1.9) serves as the
first step in the biosynthesis of cholesterol and other isoprenoid compounds in humans
(see section 1.3A for a reminder of what isoprenoids are). First, a transthioesterase
reaction transfers the acetyl group of the first acetyl CoA to a cysteine side chain in the
enzyme's active site (steps a, b). (This preliminary event is typical of many enzymecatalyzed Claisen condensation reactions, and serves to link the electrophilic substrate
covalently to the active site of the enzyme).
188
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
In the 'main' part of the Claisen condensation mechanism, the α-carbon of a second acetyl
CoA is deprotonated (step 1), forming a nucleophilic enolate.
The second acetyl CoA substrate loses a
proton to become an enolate nucleophile
First, an acetyl group is transferred to an active
site cysteine (see section 11.x for mechanism)
O
HSCoA
O
H3C
SCoA
O
O
enz
H3C
steps a,b
S
SCoA
step 1
SCoA
H
B:
SH
enz
step 2
O
H3C
O
O
SCoA
S
step 3
H3C
SCoA
SH
enz
A
O
H
enz
fig 34
The enolate carbon attacks the electrophilic thioester carbon, forming a tetrahedral
intermediate (step 2) which collapses to expel the cysteine thiol (step 3).
Exercise 13.4: Draw a mechanism for this condensation reaction between two fatty acylthioester substrates. R1 and R2 can be hydrocarbon chains of various lengths. (J. Biol.
Chem. 2011, 286, 10930.)
O
+
R2
SCoA
R1
O
HSCoA
O
S
enz
O
R2
S
R1
enz
fig 34a
In an alternative mechanism, Claisen condensations in biology are often initiated by
decarboxylation at the α−carbon of a thioester, rather than by deprotonation:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
189
Chapter 13: α-carbon part II
Decarboxylation/Claisen condensation:
CO2
RSH
O
O
O
O
+
R
SR
O
step 3
SR
O
R
SR
Mechanism:
enolate nucleophile is generated by
decarboxylation (instead of deprotonation)
O
O
CO2
O
SR
O
SR
step 1
O
R
SR
step 2
O
R
RSH
O
SR
SR
step 3
O
R
O
SR
H A
fig 45d
The thing to notice here is that the nucleophilic enolate (in red) is formed in the first step
by decarboxylation, rather than by deprotonation of an α-carbon. Other than that, the
reaction looks just like the Claisen condensation reactions we saw earlier.
Now, we can finally understand the fatty acid chain-elongation step that we heard about
in the chapter introduction in the context of the Lorenzo's oil story, which is a
decarboxylation/Claisen condensation between malonyl-ACP (the donor of a two-carbon
unit) and a growing fatty acyl CoA molecule. Notice that, again, the electrophilic acyl
group is first transferred to an active site cysteine, which then serves as the leaving group
in the carbon-carbon bond forming process.
190
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
Chain elongating (Claisen condensation) reaction in fatty acid biosynthesis
(step 1 in fatty acid synthesis cycle)
O
R
CO2
HSACP
O
+
SACP
O2C
growing fatty acid
O
O
R
SACP
SACP
malonyl-ACP
Mechanism:
O
O
R
O
SACP
SACP
O
O
step 1
SACP
malonyl ACP
step 2
H A
O
R
SACP
O
SACP
O
R
O
2
step 3
1
SACP
fatty acid chain has
grown by two carbons
fig 45b
Exercise 13.5: Curcumin is the compound that is primarily responsible for the distinctive
yellow color of turmeric, a spice used widely in Indian cuisine. The figure below shows
the final step in the biosynthesis of curcumin. Draw a mechanism for this step.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
191
Chapter 13: α-carbon part II
O
O
enz
H3CO
O2C
+
S
OCH3
OH
HO
CO2
O
O
H3CO
OCH3
curcumin
HO
OH
13.3C: Retro-Claisen cleavage
Just like the aldol mechanism, Claisen condensation reactions often proceed in the 'retro',
bond-breaking direction in metabolic pathways.
A typical Retro-Claisen cleavage reaction
(thiol nucleophile)
β−ketothioester
O
RS
C
O
C
C
O
+
R
RSH
RS
C
O
+
CH3
RS
C
H H
192
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
R
Chapter 13: α-carbon part II
Mechanism:
O
RS
RS
O
O
C
C
C
carbon-carbon
bond is broken
RS
R
step 1
H H
H SR
O
RS
C
R
H A
O
O
C R
C H
H
step 2
RS
C
C
H
H
enolate leaving
group
step 3
:B
O
RS
C
CH3
fig 35
In a typical retro-Claisen reaction, a thiol (or other nucleophile such as water) attacks the
carbonyl group of a β-thioester substrate (step 1), and then the resulting tetrahedral
intermediate collapses to expel an enolate leaving group (step 2) - this is the key carboncarbon bond-breaking step. The leaving enolate reprotonates (step 3) to bring us back to
where we started, with two separate thioesters. You should look back at the general
mechanism for a forward Claisen condensation and convince yourself that the retroClaisen mechanism illustrated aboveis a step-by-step reverse process.
Exercise 13.6: Is a decarboxylation/Claisen condensation step also likely to be
metabolically relevant in the 'retro' direction? Explain.
When your body 'burns' fat to get energy, it is a retro-Claisen cleavage reaction (EC
2.3.1.16) that is responsible for breaking the carbon-carbon bonds in step IV of the fatty
acid degradative pathway. A cysteine thiol on the enzyme serves as the incoming
nucleophile (step 1 in the mechanism below), driving off the enolate leaving group as the
tetrahedral intermediate collapses (step 2). The enolate is then protonated to become
acetyl CoA (step 3), which goes on to enter the citric acid (Krebs) cycle. Meanwhile, a
transthioesterification reaction occurs (steps a and b) to free the enzyme's cysteine
residue, regenerating a fatty acyl CoA molecule which is two carbons shorter than the
starting substrate.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
193
Chapter 13: α-carbon part II
The retro-Claisen reaction (step IV) in fatty acid degradation
O
O
HSCoA
O
R
O
R
SCoA
H3C
+
SCoA
SCoA
acetyl-CoA
fatty acid thioester is shorter
by two carbons
Mechanism:
O
O
R
O
SCoA
H
B:
step 1
R
O
O
SCoA
S
O
R
step 2
+
S
SCoA
enz
enz
S
H A
HSCoA
steps a, b
enz
fatty acid has lost two
carbons (C1 and C2 of
acetyl CoA)
step 3
O
R
SCoA
H
O
+
1
2
SCoA
acetyl CoA
S
enz
fig 36
Exercise 13.7: In a step in the degradation if the amino acid isoleucine, the intermediate
compound 2-methyl-3-keto-butyryl CoA undergoes a retro-Claisen cleavage. Predict the
products..
O
HSCoA
O
SCoA
fig 36a
Exercise 13.8: Many biochemical retro-Claisen steps are hydrolytic, meaning that water,
rather than a thiol as in the example above, is the incoming nucleophile that cleaves the
carbon-carbon bond. One example (EC 3.7.1.2) occurs in the degradation pathway for
tyrosine and phenylalanine:
194
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
H2O
O2C
O2C
CO2
O
O
CO2
+
O
O
O
fig 36a
a) Propose a likely mechanism for the reaction shown.
b) The β-diketone substrate in the reaction above could hypothetically undergo a different
retro-Claisen cleavage reaction in which the nucleophilic water attacks the other ketone
group. Predict the products of this hypothetical reaction.
Section 13.4: Conjugate addition and β-elimination
In this section, we will look at two more common biochemical reactions that proceed
through enolate intermediates. In a typical conjugate addition, a nucleophile and a
proton are 'added' to the two carbons of an alkene which is conjugated to a carbonyl (i.e.
in the α−β position). In an β-elimination step, the reverse process occurs:
Nu
H
+
R
β
α
Nu
conjugate
addition
O
R
R
β-elimination
β
O
α
R
H
fig 46
In chapter 9 we learned about nucleophilic carbonyl addition reactions, including the
formation of hemiacetals, hemiketals, and imines. In all of these reactions, a nucleophile
directly attacks a carbonyl carbon.
H A
O
R1
C
OH
R2
R1
:Nu
C
R2
Nu
fig 47
If, however, the electrophilic carbonyl is β-unsaturated - if, in other words, it contains a
double bond conjugated to the carbonyl - a different reaction pathway is possible. A
resonance structure can be drawn in which the β-carbon has a positive charge, meaning
that the β-carbon also has the potential to be an electrophilic target.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
195
Chapter 13: α-carbon part II
β-carbon is electrophilic
O
O
R β
R β
R
α
R
α
fig 48
If a nucleophile attacks at the β-carbon, an enol or enolate intermediate results (step 1
below). In many cases this intermediate collapses and the α-carbon is protonated (step
2). This type of reaction is known as a conjugate addition.
Mechanism of a conjugate addition reaction
:B
Nu
H
O
R β
α
Nu
R
step 1
R β
Nu
O
α
R β
step 2
R
α
O
R
H
H A
enolate
fig 49
The reverse of a conjugate addition is a β-elimination, and is referred to mechanistically
the abbreviation E1cb.
Mechanism of an E1cB elimination
X
R
β
H A
X
O
α
R
step 1
R
β
O
α
O
R
step 2
R
β
α
R
+
X
H
H
:B
fig 50
The E stands for 'elimination'; the numeral 1 refers to the fact that, like the SN1
mechanism, it is a stepwise reaction with first order kinetics. The 'cB' designation refers
to the intermediate, which is the conjugate base of the starting compound. In step 1, an
α-carbon is deprotonated to produce an enolate, just like in aldol and Claisen reactions
we have already seen. In step 2, the excess electron density on the enolate expels a
leaving group at the β position (designated 'X' in the figure above). Notice that the α and
β carbons change from sp3 to sp2 hybridization with the formation of a conjugated double
bond.
196
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
(In chapter 14 we will learn about alternate mechanisms for alkene addition and βelimination reactions in which there is not an adjacent carbonyl (or imine) group, and in
which the key intermediate species is a resonance-stabilized carbocation. )
Step II of fatty acid degradation is a conjugate addition of water, or hydration.
B:
H
H O
OH
O
R
R
SCoA
OH
O
R (S)
H H
SCoA
H
H
O
H A
SCoA
fig 51
Note the specific stereochemical outcome: in the active site, the nucleophilic water is
bound behind the plane of the conjugated system (as drawn in the figure above), and the
result is S configuration in the β-hydroxy thioester product.
In step III of the fatty acid synthesis cycle we saw an E1cb β-elimination of water
(dehydration):
OH
H A
OH
O
R (R)
H H
SACP
O
R
O
SACP
R
SACP
H
+
:B
HOH
fig 52
Notice that the stereochemistry at the β-carbon of the starting alcohol is R, whereas the
hydration pathway (step II) reaction in the fatty acid degradation cycle pathway results in
the S stereoisomer. These two reactions are not the reverse of one another!
Here is another example of a β-elimination reaction (EC 4.2.3.4), this time with a
phosphate leaving group. It is part of the biosynthesis of aromatic amino acids.
B:
H
HO CO2
HO CO2
HO CO2
O
O
O
α
OH
β
OP
O
OH
OP
+ Pi
H2C
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OH
O
197
Chapter 13: α-carbon part II
fig 53
Exercise 13.9: In the glycolysis pathway, the enzyme 'enolase' (EC 4.2.1.11) catalyzes the
E1cb dehydration of 2-phosphoglycerate. Predict the product of this enzymatic step.
H2O
OP
O
OH
O
2-phosphoglycerate
Exercise13.10: N-ethylmaleimide (NEM) is an irreversible inhibitor of many enzymes that
contain active site cysteine residues. Show how this inactivation could occur through a
conjugate addition of cysteine, and show the structure of the labeled residue. (Michael
addition)
O
N
O
NEM
fig 54
Exercise 13.11: Argininosuccinate lyase (4.3.2.1), an enzyme in the metabolic pathway
that serves to eliminate nitrogen from your body in the form of urea in urine, catalyzes
this β-elimination step:
NH2
O2C
NH3
N
N
H
H
CO2
NH2
CO2
O2C
N
NH3
H
arginine
arginosuccinate
+
O2C
CO2
fumarate
fig 54
198
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
NH2
Chapter 13: α-carbon part II
Propose a complete mechanism. Hint: Don't be intimidated by the size or complexity of
the substrate - review the β-elimination mechanism, then identify the leaving group and
breaking bond, the α-carbon which loses a proton, the carbonyl that serves to stabilize the
negatively-charged (enolate) intermediate, and the double bond that forms as a result of
the elimination. You may want to designate an appropriate 'R' group to reduce the
amount of drawing.
Section 13.5: Carboxylation by the Rubisco enzyme
It is difficult to overstate the importance to biology and ecology of the enzymatic reaction
we are going to see next: ribulose 1,5-bisphosphate carboxylase (Rubisco) plays a key
role in closing the 'carbon cycle' in our biosphere, catalyzing the incorporation of a
carbon atom - in the form of carbon dioxide from the atmosphere - into organic
metabolites and eventually into carbohydrates, lipids, nucleic acids, and all of the other
organic molecules in living things. Rubisco is probably the most abundant enzyme on
the planet.
You can think of a carboxylation reaction as essentially a special kind of aldol reaction,
except that the carbonyl electrophile being attacked by the enolate is CO2 rather than a
ketone or aldehyde:
Mechanism for carboxylation of an enolate
O
R
C
O
O
CH2
enolate
C
R
C
O
O
C
C
O
H H
fig 38
Here is the full Rubisco reaction. Notice that the carbon dioxide (in blue) becomes
incorporated into one of the two phosphoglycerate products.
The Rubisco reaction
CO2
H2O
OH
OH
O
PO
OP
OH
PO
O
O
HO H
ribulose-1,5-bisphosphate
+
O
OP
O
2x phosphoglycerate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
199
Chapter 13: α-carbon part II
Mechanism:
H A
OH
PO
5
4
O
3
OH
OP
2
1
PO
3
step 1
HO H
OH
O
2
step 2
O
:B
H
ribulose-1,5-bisphosphate
PO
OP
OH
PO
5
4
3
O
O
+
O
2
O
O
C
:B
O
H2O
OP
1
OP
2
O
carboxylation
OH
OH
3
OH HO CO2
PO
steps 4, 5, 6
(hydrolytic retro-Claisen)
step 3
3
2
OP
O
2x phosphoglycerate
fig 37 fig 39
The mechanism for the Rubisco reaction is somewhat involved, but if we break it down
into its individual steps, it is not terribly difficult to follow. In step 1, an α-carbon on
ribulose 1,5-bisphosphate is deprotonated to form an enolate. In step 2, the oxygen at
carbon #3 is deprotonated while the oxygen at carbon #2 is protonated: combined, these
two steps have the effect of creating a different enolate intermediate and making carbon
#2, rather than carbon #3, into the nucleophile for an aldol-like addition to CO2 (step 3).
Carbon dioxide has now been 'fixed' into organic form - it has become a carboxylate
group on a six-carbon sugar derivative. Steps 4, 5, and 6 make up a hydrolytic retroClaisen cleavage reaction (in other words, water is the bond-breaking nucleophile)
producing two molecules of 3-phosphoglycerate. Phosphoglycerate is channeled into the
gluconeogenesis pathway to eventually become glucose.
Exercise 13.12: Draw out the full mechanism for steps 4-6 in the Rubisco reaction.
200
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
Key learning objectives for this chapter
Before moving on to the next chapter, you should:
Be able to draw reasonable mechanisms for reactions of the following type:
Decarboxylation of a β-carboxy ketone or aldehyde
Claisen condensation and retro-Claisen cleavage
Hybrid decarboxylation-Claisen condensation
Conjugate addition
E1cb elimination
Understand (though not necessarily memorize) the fatty acid synthesis and degradation
cycles, and how the Claisen. retro-Claisen, conjugate addition, and E1cb elimination
steps fit in.
Be able to draw a complete mechanism for the Rubisco reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
201
Chapter 13: α-carbon part II
Problems
P13.1: Tetrahydrolipastatin, a potent inhibitor of lipase enzymes (see section 11.6) is
being tested as a possible anti-obesity drug. Lipastatin, a close derivative, is synthesized
by the bacterium Streptomyces toxytricini. The biosynthetic pathway involves the
following step shown below - draw a likely mechanism. (J. Biol. Chem. 1997, 272, 867).
OH
O
SCoA
OH
+
O
SCoA
O
O
SCoA
P13.2: The metabolism of camphor by some bacteria involves the step below. Draw a
likely mechanism. (J. Biol. Chem 2004, 279, 31312)
H2O
O
O2C
O
O
P13.3: The glucogeogenesis pathway, by which glucose is synthesized from pyruvate,
begins with a reaction catalyzed by pyruvate carboxylase. The enzyme requires the CO2carrying biotin to function, but the final step is thought to be the simple carboxylation of
pyruvate by free carbon dioxide (Biochem. J. 2008, 413, 369). Draw a mechanism for this
step.
202
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
O
CO2
O
O
O
O
O
O
O
oxaloacetate
pyruvate
P13.4: Draw a reasonable mechanism for this decarboxylation step in tryptophan
biosynthesis (EC 4.1.1.45). Hint: a tautomerization step precedes the decarboxylation.
O
NH3
O
CO2
O
O
O
NH3
O
O
O
P13.5: The biosynthetic pathway for the antibiotic compound rabelomycin begins with
the condensation of malonyl CoA and acetyl CoA. Predict the product of this reaction,
and propose a likely mechanism. (Org. Lett. 2010 12, 2814.)
O
O
O
CO2
HSCoA
O
+
SCoA
CoAS
malonyl CoA
CH3
acetyl CoA
P13.6 Predict the product of this decarboxylation step in the biosynthesis of the amino
acid tyrosine. Hint: think about comparative stability when you are considering where
protonation will occur!
CO2
H+
CO2
CO2
O
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
203
Chapter 13: α-carbon part II
P13.7: Show a likely mechanism for this reaction from lysine biosynthesis:
OH
H2O
O2C
N
CO2
N
O2C
CO2
P13.8: Compound A undergoes hydrolytic cleavage in some fungi to form the products
shown. Predict the structure of A. (J. Biol. Chem. 2007, 282, 9581)
H2O
O
A
O
O
CH3
+
O
O
O
P13.9: Propose a mechanism for the following reaction from the gluconeogenesis
pathway (EC 4.1.1.32):
GTP
CO2
GDP
OP
O
CO2
O2C
O2C
C
CH2
P13.10: Dehydroquinate undergoes dehydration (EC 4.2.1.10) in aromatic amino acid
biosynthesis. Experimental and genomic evidence points to a lysine-linked iminium
intermediate. More than one dehydration product is possible for dehydroquinate, but in
this case the most stable product is the one that forms. Predict the structure of the
product, explain why it is the more stable of the possible dehydration products, and draw
a mechanism for its formation.
HO CO2
H2O
NH3
enz
O
OH
OH
dehydroquinate
P13.11: The enzyme catalyzing the reaction below, thought to participate in the
fermentation of lysine in bacteria, was recently identified and characterized (J. Biol.
Chem. 2011, 286, 27399). Propose a likely mechanism. Hint: the mechanism involves
204
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
two separate carbon-carbon bond forming and bond breaking steps. C1 of acetyl CoA is
identified with a red dot to help you trace it through to the product.
NH3
O
O
CO2
NH3
+
H3C
O
O
+
SCoA
CO2
SCoA
P13.12: Menaquinone (Vitamin K) biosynthesis in bacteria includes the following step:
O
OH
H2O
O
O
O
SCoA
SCoA
O
OH
Propose a likely mechanism. Hint: the mechanism involves a Claisen condensation step
which is unusual in that the electrophile is a carboxylic acid group rather than a thioester.
What is the driving force that allows this unusual step to occur? (J. Biol. Chem. 2010,
285, 30159)
P13.13: 4-maleylacetoacetate isomerase (EC 5.2.1.2) catalyzes the following cis to trans
alkene isomerization as part of the degradation of the aromatic amino acids phenylalanine
and tyrosine.
CO2 O
O
O
CO2
O
CO2
O2C
4-fumarylacetoacetate
4-maleylacetoacetate
NH3
H
N
O2C
O
N
CO2
H
O
HS
Glutathione
(GSH)
fig5
The enzyme uses the thiol-containing coenzyme glutathione, which is also involved in
the formation of disulfide bonds in proteins, but in this case glutathione serves as a 'thiol
group for hire'. The mechanism for the reaction is essentially a reversible conjugate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
205
Chapter 13: α-carbon part II
addition of glutathione. Draw out the steps for this mechanism, showing how the cistrans isomerization could be accomplished. Also, explain why the equilibrium for this
reaction favors 4-fumarylacetoacetate. The structure of glutathione is shown, but you can
use the abbreviation GSH in your mechanism.
P13.14: Based on the mechanistic patterns you have studied in this chapter, propose a
likely mechanism for this final reaction in the degradation of the amino acid cysteine in
mammals.
O
O
O
O
S
O
O
O
H3C
S
+
O
O
O
pyruvate
P13.15: Propose a mechanism for the following carboxylation reaction (EC 6.4.1.4) in
the leucine degradation pathway. The complete reaction is dependent on the CO2carrying coenzyme biotin as well as ATP, but assume in your mechanism that the actual
carboxylation step occurs with free CO2 (you don’t need to account for the role played by
biotin or ATP).58, bottom)
O
O
+
O
CO2
SCoA
O
SCoA
P13.16: Propose a mechanism for the following reaction, which is part of the degradation
pathway for the nucleotide uridine.
(
O
NH
N
H
2 H2O
O
NH4+
CO2
O
O
NH2
P13.17: Illustrated below is a series of reactions in the degradation pathway for the
amino acid methionine. In step 1, an alcohol group on C3 is oxidized to a ketone, and in
step 4 the ketone is reduced back to an alcohol - we will study these reactions in chapter
X. In steps 2 and 3, the thiol (homocysteine) is replaced by water - but this does NOT
involve a nucleophilic substitution process.
a) Draw a likely mechanism for step 2
206
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
b) Draw a likely mechanism for step 3
c) How does the involvement of the redox steps (steps 1 and 4) provide evidence that
overall substitution of water for homocysteine is not a nucleophilic substitution?
NH3
(A = Adenine)
=R
5
O2C
S
H
4
HO
H
O
A
1
3
2
oxidation
step 1
O
A
RS
O
OH
OH
H2O
steps 2, 3
OH
OH
O
A
O
reduction
step 4
HO
OH
O
A
OH
adenosine
+
NH3
O2C
SH
homocysteine
P13.18: (Challenging!) A recently discovered reaction in the biosynthesis of rhizoxin, a
potent virulence factor in the rice-seedling blight fungus Rhizopus microsporus, is
illustrated below (Angewandte Chemie 2009, 48, 5001). The reaction takes place at the
intersection of two 'modules' of a multi-enzyme complex, and provides an example of a
biochemical conjugate addition step that results in the formation of a new carbon-carbon
bond (conjugate addition of a carbon nucleophile is referred to as a Michael addition).
Draw a likely mechanism.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
207
Chapter 13: α-carbon part II
O
OH
enz1
O
R
S
enz2
+
O
S
O
CO2
O
O
O
enz1
R
S
P13.19: Suggest a mechanism for the following transformation. (Hint – only two
mechanistic steps are required.)
HO CO2
HO CO2
O
H2C
O
OH
OH
OH
OH
P13.20:
a) The 'acetoacetic ester synthesis' is a useful way to synthesize methyl esters in the lab.
The reaction mechanism is described as α-carbon deprotonation to form an enolate,
followed by SN2 alkylation, ester hydrolysis, and decarboxylation. Below is an example:
O
O
O
CH3O-
1-bromopropane
HCl, H2O
heat
OCH3
Draw out a reasonable mechanism, taking care to propose reactive intermediates that are
appropriate given the basic or acidic conditions present (note that the reaction starts under
basic conditions, then is later acidified).
b) Suggest starting compounds for the synthesis of 4-phenyl-2-butanone by the
acetoacetic ester reaction.
c) A very useful ring-forming reaction in laboratory synthesis is called 'Robinson
annulation' (Sir Robert Robinson was an English chemist who won the 1947 Nobel Prize
208
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 13: α-carbon part II
in Chemistry, and the term ‘annulation’ comes from the Latin annulus, meaning ‘ring’.)
The reaction, which takes place in basic conditions, consists of a conjugate (Michael)
addition step, followed by aldol addition and finally dehydration (β-elimination of water).
A typical example is shown below, with carbons numbered to help you to follow the
course of the reaction.
1
7
6
O 5
+
2
3
4
CH3
O
1
7
OH-
6
O 5
CH3
2
3
4
Draw a mechanism for this reaction (when proposing intermediate species, keep in mind
that the reaction is occurring in a basic environment, and choose protonation states
accordingly).
d) Propose starting compounds for the Robinson annulation synthesis of the following
product:
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
209
Chapter 14: Electrophilic reactions
Chapter 14
Electrophilic reactions
Introduction
Linnda Caporael probably should have paid a little more attention to the graduation
requirements in her college catalog. Going through the graduation checklist during her
senior year, she discovered to her dismay that she still needed to fulfill a social science
requirement, so she promptly enrolled in an American History course. It was a decision
that would in time lead to her authoring a paper in a prestigious scientific journal, being
featured in a front page story in the New York Times, and changing our understanding of
one of the most intriguing - and disturbing – episodes in American history.
Professor Caporael (Linnda went on to become a professor of Behavioral Psychology at
Rensselaer Polytechnic Institute) recounted her story in an episode of the PBS
documentary series Secrets of the Dead. Early in the semester, she learned that as part of
her history course she would be required to complete a research paper on a topic of her
own choosing. She had recently seen a performance of The Crucible, Arthur Miller's
classic play about the Salem witch trials, and decided to do her research on Anne Putnam,
one of the young Salem girls who accused several village women of bewitching them.
The symptoms of 'bewitchment' that afflicted the girls were truly frightening: thrashing
and convulsions, visions of snakes and ferocious beasts, a sudden inability to speak, and a
feeling that ants were crawling under their skin.
These children were bitten and pinched by invisible agents: their arms, necks and
backs turned this way and that way, and returned back again, so as it was,
impossible for them to do of themselves, and beyond the power of any epileptick
fits, or natural disease to effect. Sometimes they were taken dumb, their mouth
flopped, their throats choaked, their limbs wracked and tormented. . .
210
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
(From “A Modest Enquiry Into the Nature of Witchcraft” (1702) by Reverend John Hale.
http://historyofmassachusetts.org/betty-parris-first-afflicted-girl-of-the-salem-witchtrials/)
As Linnda continued to read accounts of the 'fits' afflicting the Salem girls, she was
suddenly struck by the similarities to another, more recent episode that she had read
about. In the summer of 1951, in the French village of Pont Saint Esprit, a number of
local people were simultaneously seized by hallucinations, convulsions, and other
symptoms very much like those described during the Salem witch trials hundreds of years
earlier. Leon Armunier, a former postman in Pont Saint Esprit, described his experience
to the BBC:
It was terrible. I had the sensation of shrinking and shrinking, and the fire and the
serpents coiling around my arms . . .Some of my friends tried to get out of the
window. They were thrashing wildly. . . screaming, and the sound of the metal
beds and the jumping up and down... the noise was terrible. I'd prefer to die
rather than go through that again.
There have been several explanations offered for what caused the Pont Saint Esprit
outbreak (including that the CIA was experimenting with mass LSD poisoning as a form
of chemical warfare), but the most widely accepted theory is that the hallucinations were
caused by eating bread made from contaminated grain.
Claviceps purpurea, a fungus known to grow in rye and other grains, produces a class of
hallucinogenic compounds called 'ergot alkaloids' which are derived from lysergic acid
(the hallucinogenic drug LSD is a synthetic derivative of lysergic acid). Claviceps thrives
in damp grain, and special care must be taken to avoid contamination when storing grain
grown during warm, rainy summers.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
211
Chapter 14: Electrophilic reactions
H
HO
H3C O
O
N
CH3
O
O
N
HN
N
OH
N
N
CH3
O
N
CH3
H
H
H
N
H
N
H
N
H
(+)-lysergic acid
(+)-lysergic acid diethylamide
(LSD)
ergotamine
(an ergot alkaloid)
fig 1a
Digging deeper into the records of the Salem witchcraft episode, Linnda Caporael learned
that the summer of 1691 was unusually damp. The first cases of 'bewitchment' in Salem
village occurred in the winter of 1691-1692, when the villagers would have been
consuming grain stored from the previous summer. Rye, the kind of grain most
vulnerable to Claviceps contamination, was the staple crop in Salem at the time.
Furthermore, nearly all of the affected girls lived on farms on the swampy western edge
of the town, where Claviceps contamination would have been most likely to occur.
This was all circumstantial evidence, to be sure, but it was enough to convince Linnda
that ergot poisoning was much more plausible as a root cause of the behavior of the
afflicted girls than simply mass hysteria, which had long been the accepted theory. She
summarized her findings in a paper that was later published in the journal Science, with
the colorful title "Ergotism: The Satan Loosed in Salem?" (Science 1976, 192, 21). Her
theory is still not universally accepted, but scientists and historians are for the most part
in agreement that ergot poisoning was the cause of other outbreaks of convulsions and
hallucinations, often called 'Saint Anthony's Fire', that have occurred throughout
European history. It is possible that ergot poisoning may have played a role in literature
as well: professor Caporael, in an interview with PBS, recounts how she was recently
contacted by an historian with in intriguing idea. Is it possible that Caliban, the wild, halfhuman character in Shakespeare's The Tempest who is tormented by hallucinations
inflicted upon him by the wizard Prospero, could be a literary manifestation of ergot
poisoning episodes that occurred in England during Shakespeare's time?
For every trifle are they set upon me;
Sometime like apes that mow and chatter at me
And after bite me, then like hedgehogs which
Lie tumbling in my barefoot way and mount
Their pricks at my footfall; sometime am I
212
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
All wound with adders who with cloven tongues
Do hiss me into madness.
(William Shakespeare's The Tempest, Act II scene ii)
In this chapter, we will learn about a class of organic reaction that is central to the
biosynthesis of ergot alkaloids in Claviceps. The key first step in the biosynthetic
pathway is a reaction unlike any we have yet seen:
CO2
CO2
NH3
NH3
PPi
+
N
H
tryptophan
OPP
DMAPP
N
H
dimethylallyl tryptophan
(several steps)
OH
O
N
CH3
H
N
H
(+)-lysergic acid
fig 1b
As you can see, the first step is condensation between the amino acid tryptophan and
dimethylallyl diphosphate (DMAPP), the building block molecule for isoprenoids
(section 1.3A). What you should also notice in the reaction figure above is that a new
carbon-carbon bond is formed, and yet the chemistry involved is clearly very different
from the carbon-carbon bonding forming aldol additions and Claisen condensations we
learned about in the previous two chapters: there is no carbonyl to be found anywhere
near the site of reaction, and one of the bond-forming carbons is part of an aromatic ring.
We will see later in this chapter that this reaction mechanism is classified as an
'electrophilic aromatic substitution', and is one of a broader family of organic reaction
mechanisms that includes electrophilic additions, substitutions, and isomerizations.
'Electrophilic' is the key term here: in organic chemistry, an 'electrophilic' reaction
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
213
Chapter 14: Electrophilic reactions
mechanism is one in which the π-bonded electrons in a carbon-carbon double (or
sometimes triple) bond are drawn towards an electron-poor species, often an acidic
proton or carbocation. In essence, the π bond is acting as a nucleophile or base.
π electrons drawn toward proton, actiing as base
π electrons drawn toward electrophile, actiing as nucleophile
H A
E
H
E
or
C C
C C
C C
C C
fig 1
Notice above that, once the π bond breaks and a new s bond forms, the second carbon
that was part of the original π bond becomes a carbocation. Carbocation intermediates
play a critical role in this chapter, because a carbocation is a highly reactive species and
will quickly attract a pair of electrons. The stability of the carbocation intermediate
(recall that we learned about carbocation stability in section 8.5), and the manner in
which it accepts a pair of electrons, plays a key role in determining the outcome of the
reaction.
Section 14.1: Electrophilic addition
14.1A: Addition of HBr to alkenes
The simplest type of electrophilic reaction to visualize is the addition of a haloacid such
as HBr to an isolated alkene. It is not a biological reaction, but nonetheless can serve as a
convenient model to introduce some of the most important ideas about electrophilic
reactions.
Electrophilic addition of HBr to an alkene:
Br
H Br
R
R
C C
R
R
R
step 1
H
C C R
R
R
Br H
step 2
R C C R
R R
carbocation intermediate
fig 1
Step 1 is an acid-base reaction: the π electrons of the alkene act as a base and extract the
acidic proton of HBr. This leaves one of the carbons with a new bond to hydrogen, and
the other with an incomplete octet and a positive formal charge. In step 2, the
nucleophilic bromide anion attacks the electrophilic carbocation to form a new carbon214
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
bromine bond. Overall, the HBr molecule - in the form of a proton and a bromide anion has been added to the double bond.
To understand how π-bonded electrons in an alkene could be basic, let's first review the
bonding picture for alkenes. Recall (section 2.1C) that the both of the carbons in an
alkene group are sp2 hybridized, meaning that each carbon has three sp2 hybrid orbitals
extending out in the same plane at 180o angles (trigonal planar geometry), and a single,
unhybridized p orbital oriented perpendicular to that plane - one lobe above the plane,
one lobe below.
'loose' electron
density in π bond
fig 2
The unhybridized p orbitals on the two alkene carbons overlap, in a side-by-side fashion,
to form the π bond, which extends above and below the plane formed by the σ bonds.
The two electrons shared in this π bond are farther away from the carbon nuclei than the
electrons in the carbon-carbon σ bond, and thus are more accessible to the acidic proton.
In addition, recall that molecular orbital (MO) theory tells us that π orbitals are higher in
energy than σ orbitals (section 2.2). As a consequence, it is easier to break the π bond of
an alkene than it is to break the σ bond: the π bond is more reactive.
As the HBr molecule approaches the alkene, a new σ bond is formed between one of the
alkene carbons and the electron-poor proton from HBr. The carbon, which was sp2
hybridized when it was part of the alkene, is now sp3 hybridized. The other alkene
carbon is still sp2 hybridized, but it now bears a positive formal charge because it has
only three bonds, and its p orbital is empty. But it won't stay empty for long: a
carbocation is a very reactive, unstable intermediate. The bromide ion will rapidly act as
a nucleophile, filling the orbital with a pair of electrons, and now with four σ bonds the
carbon is sp3-hybridized.
The first step in the electrophilic addition reaction is much slower than the second step,
because the intermediate carbocation species is higher in energy than either the reactants
or the products, and as a result the energy barrier for the first step is also higher than for
the second step. The slower first step is the rate-determining step: a change in the rate of
the slow step will effect the rate of the overall reaction, while a change in the rate of the
fast step will not.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
215
Chapter 14: Electrophilic reactions
transition state 1
transition state 2
energy barrier is low: faster step
energy barrier is high: slower step
R
H
C C R
R
R
Br
intermediates
R
R
B
C C
R
R
Br H
H Br
R C C R
reactants
R R
product
(fig 3)
fig 2a
It is important to recognize the inherent difference between an electrophilic addition to an
alkene and a conjugate addition to an alkene in the a-b position, the latter of which we
studied earlier in section 13.4. In both reactions, a proton and a nucleophile add to the
double bond of an alkene. In a conjugate addition, the nucleophilic attack takes place
first, resulting in a negatively charged intermediate (an enolate). Protonation is the
second step. Also, of course, the alkene must be conjugated to a carbonyl or imine.
conjugate addition:
protonation second
Nu
H A
Nu
H
Nu
O
O
O
negatively charged intermediate
electrophilic addition:
H Nu
protonation first
Nu
H
H
Nu
positively charged intermediate
fig 3
In an electrophilic addition, proton abstraction occurs first, generating a positivelycharged intermediate. Nucleophilic attack is the second step. No conjugated carbonyl or
imine group is required: in fact a nearby carbonyl group would actually slow down a
216
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
hypothetical electrophilic addition reaction down because a carbonyl is an electron
withdrawing, carbocation-destabilized group.
14.1B:The stereochemistry of electrophilic addition
Depending on the structure of the starting alkene, electrophilic addition has the potential
to create two new chiral centers. Addition of HBr to an alkene is not stereoselective: the
reaction results in racemization at both of the alkene carbons. Consider the addition of
HBr to cis-3,4-dimethyl-3-hexene. The initial proton abstraction step creates a new chiral
center, and because the acidic proton could be added to either side of the planar alkene
carbon with equal probability, the center could have either S or R configuration.
H
Br
R
R
Br- in front
Br
H
syn addition
R
H
Br- in back
H Br
Br
S
R
HBr in front
anti addition
H
Br
HBr in back
R
Br- in front
Br
S
anti addition
H
S
H
Br- in back
Br
S
S
syn addition
fig 4
Likewise, in the second step the nucleophilic bromide ion could attack from either side of
the planar carbocation, leading to an equal mixture of S and R configuration at that
carbon as well. Therefore, we expect the product mixture to consist of equal amounts of
four different stereoisomers.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
217
Chapter 14: Electrophilic reactions
Exercise 14.1: Predict the product(s) of electrophilic addition of HBr to the following
alkenes. Draw all possible stereoisomers that could form, and take care not to draw
identical structures twice.
a) trans-2-butene
b) cis-3-hexene
c) cyclopentene
d) cis-4,5-dimethylcyclohexene
14.1C: The regiochemistry of electrophilic addition
In many cases of electrophilic addition to an alkene, regiochemistry comes into play: the
reaction can result in the formation of two different constitutional isomers. Consider the
electrophilic addition of HBr to 2-methylpropene:
(fig 4)
Br
Br:
product A
H Br
H
a
H
Ia
3
2
1
b
H
:Br
H
product B
Ib
Br
fig 5
Note that carbon #1 and carbon #2 in the starting alkene are not the same - carbon #2 is
bonded to two methyl groups, and carbon #1 to two hydrogen atoms. The initial
protonation step could therefore go two different ways, resulting in two different
carbocation intermediates. Notice how pathway 'a' gives a tertiary carbocation
intermediate (Ia), while pathway 'b' gives a primary carbocation intermediate (Ib) We
know from section 8.5 that the tertiary carbocation Ia is lower in energy. Consequently,
the transition state TS(a) leading to Ia is lower in energy than TS(b), meaning that Ia
forms faster than Ib.
218
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
TSb
IA is more stable,
thus TSa is lower
TSa
Ib:
Ib
H3C H
CH3
Ia:
Ia
(forms faster)
H
A
H3C Br
H
B
H3C H
Br
fig 6
Because the protonation step is the rate determining step for the reaction, tertiary alkyl
bromide A will form faster than the primary alkyl bromide B, and thus A will be the
predominant product of the reaction. The electrophilic addition of HBr to 2methylpropene is regioselective: more than one constitutional isomer can potentially
form, but one isomer is favored over the other. It is generally observed that in
electrophilic addition of haloacids to alkenes, the more substituted carbon is the one that
ends up bonded to the heteroatom of the acid, while the less substituted carbon is
protonated. This 'rule of thumb' is known as Markovnikov's rule, after the Russian
chemist Vladimir Markovnikov who proposed it in 1869.
While it is useful in many cases, Markovikov's rule does not apply to all electrophilic
addition reactions. It is better to use a more general principle:
The regioselectivity of electrophilic addition
When an asymmetrical alkene undergoes electrophilic addition, the product that
predominates is the one that results from the more stable of the two possible carbocation
intermediates.
How is this different from Markovnikov's original rule? Consider the following
hypothetical reaction, in which the starting alkene incorporates two trifluoromethyl
substituents:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
219
Chapter 14: Electrophilic reactions
lower energy
carbocation
CF3
F3C
F3C H
F3C H
F3C
F3C
Br:
H Br
X
Br
anti-Markovnikov product
higher energy
carbocation
CF3
F3C
fig 7
Now when HBr is added, it is the less substituted carbocation that forms faster in the ratedetermining protonation step, because in this intermediate the carbon bearing the positive
charge is located further away from the electron-withdrawing, cation-destabilizing
fluorines. As a result, the predominant product is the secondary rather than the tertiary
bromoalkane. This is referred to as an anti-Markovnikov addition product, because it
'breaks' Markovnikov's rule.
If the two possible carbocation intermediates in an electrophilic addition reaction are of
similar stability, the product will be a mixture of constitutional isomers.
Br
Br:
H Br
a
roughly equal mixture of constitutional
isomers - not a regiospecific reaction
b
:Br
Br
both intermediates are 3o carbocations
fig 8
14.1D: Electrophilic addition of water and alcohol
The (non-biochemical) addition of water to an alkene is very similar mechanistically to
the addition of a haloacid such as HBr or HCl, and the same stereochemical and
regiochemical principles apply. A catalytic amount of a strong acid such as phosphoric or
sulfuric acid is required, so that the acidic species in solution is actually H3O+. Note that
H3O+ is regenerated in the course of the reaction.
220
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
H2O
H OH2
H2O
CH3
H3C
C
C
CH3
H3C
H
C
C
H
H H
H
H
H3C O
C
H3C
CH3
H
H3O
H3C
H3C
OH
C
CH3
fig 9
If an alkene is treated with methanol and a catalytic amount of strong acid, the result is an
ether:
indicates that a catalytic amount of a
strong acid is used in the reaction
CH3
H3C
C
C
H
+
CH3OH
H3C
H+
H3C
OCH3
C
CH3
H
fig 10
Exercise 14.2: Draw a mechanism for the ether-forming reaction in figure 10.
14.1E: Addition to conjugated alkenes
Electrophilic addition to conjugated alkenes presents additional regiochemical
possibilities, due to resonance delocalization of the allylic carbocation intermediate.
Addition of one molar equivalent of HBr to 1,3-butadiene, for example, leads to a
mixture of three products, two of which are a pair of enantiomers due to the creation of a
chiral center at carbon #2.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
221
Chapter 14: Electrophilic reactions
H Br
:Br
Br
+
Br
Br:
Br
Exercise 14.3: Explain why 4-bromo-1-butene is not a significant product of the reaction
above.
Exercise 14.4: Predict the major product(s) of the following reactions. Draw all possible
stereoisomers, and take care not to draw the same structure twice.
a)
+
HBr
b)
+
HI
c)
F
+
H+
H2O
d) Hint - are the double bonds in an aromatic ring likely to undergo electrophilic
addition?
+
222
CH3OH
H+
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
e)
H+
H2O
+
f)
+
HCl
fig 12
14.1F: Biochemical electrophilic addition reactions
Myrcene is an isoprenoid compound synthesized by many different kinds of plants and
used in the preparation of perfumes. Recently an enzymatic pathway for the degradation
of myrcene has been identified in bacteria. The first step of this pathway is electrophilic
addition of water to a conjugated alkene system. J. Biol. Chem 2010, 285, 30436.
H2O
OH
lanalool
myrcene
fig 15
Exercise 14.5: Draw a mechanism for the above reaction.
Although the hydration of myrcene above looks very familiar, many enzyme-catalyzed
electrophilic addition reactions differ from what we have seen so far, in that the electronpoor species attacked by the π-bonded electrons in the initial step is a carbocation rather
than an acidic proton:
electrophile is carbocation, not proton
H
C
H
:B
H
R
R
O
H
step 1
R
C
OH
R
H H
step 2
R
C
R
H H
fig 13
a-terpineol, a major component in the sap of pine trees, is formed in an electrophilic
addition reaction. The first thing that happens (which we will refer to below as 'step a', in
order to keep the step numbering consistent what the addition mechanisms we have seen
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
223
Chapter 14: Electrophilic reactions
so far) is departure of a pyrophosphate leaving group, forming an allylic carbocation
electrophile.
OPP
CH3
step a
CH3
H
O H
:B
carbocation electrophile
CH3
step 1
CH3
H
π electrons attack electrophile
step 2
OH
CH3
(S)−α-terpineol
fig 14
The actual electrophilic addition stage of the reaction begins with step 1, as the π
electrons an alkene are drawn toward one of the two carbons that share the positive
charge, effectively closing a six-membered ring. A water molecule then attacks the
second carbocation intermediate (step 2), which completes the addition process.
Notice something important about the regiochemical course of the reaction: step 1 results
in the formation of a six-membered ring and a tertiary carbocation. As we have stressed
before, biochemical reactions tend to follow energetically favorable mechanistic
pathways.
Exercise 14.6: An alternate regiochemical course to step 1 shown above could result in a
seven-membered ring and a secondary carbocation, a much less energetically favorable
intermediate in terms of both carbocation stability and ring size. Draw a mechanism for
this hypothetical alternate reaction, and show the product that would result after the
addition of water in a hypothetical 'step 2'.
Section 14.2: Elimination by the E1 mechanism
14.2A: E1 elimination - an overview
The reverse of electrophilic addition is called E1 elimination. We will begin by looking
at some non-biochemical E1 reactions, as the E1 mechanisms is actually somewhat
unusual in biochemical pathways.
224
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
E1 elimination:
B:
X
H
H
R
C C
R C C R
R C C R
step 1
R R
R
step 2
R R
R
R
fig 16
An E1 elimination begins with the departure of a leaving group (designated 'X' in the
general figure above) and formation of a carbocation intermediate (step 1). Abstraction
of a proton from an adjacent carbon (step 2) sends two electrons down to fill the empty p
orbital of the carbocation, forming a new π bond. The base in this step may be the leaving
group, or another basic species in solution.
E1 elimination does not occur when the leaving group is bonded to a primary carbon,
unless the carbon is in the allylic or benzylic position. Recall that a primary carbocation,
unless stabilized by resonance, is highly unstable and an unlikely reaction intermediate.
E1 eliminations can occur at secondary carbons, however. If cyclohexanol is heated with
a catalytic amount of phosphoric acid, elimination of water (dehydration) results in
cyclohexene as the product. The role of the phosphoric acid is to protonate the alcohol
('step a' below), making it a viable leaving group.
O
poor leaving group
O P OH
H
good leaving group
OH
OH
O
OH2
O P OH
step a
H
step 1
OH
H
O
step 2
O P OH
H
OH
fig 17
The reaction is reversible, but if cyclohexene is distilled away from the reaction mixture
as it forms, the equilibrium can be driven towards product (you may want to review Le
Chatelier's principle in your General Chemistry textbook). Separation of cyclohexene
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
225
Chapter 14: Electrophilic reactions
(boiling point 83 oC) from cyclohexanol (boiling point 161 oC) is simple because of the
large difference in boiling points between the two liquids.
Exercise 14.7: When the laboratory reaction described above is run to completion, a
viscous 'goop' is usually left over in the distillation flask, which hardens upon cooling.
Draw a mechanism showing how this 'goop' might form, and explain why it hardens upon
cooling.
14.2B: Regiochemistry of E1 elimination
Nonenzymatic E1 reactions can often result in a mixture of more than one alkene
product. Elimination of 'HX' from the following starting compound, for example, could
yield three different possible alkene products.
CH3
H3C
a proton can be abstracted from one of
three adjacent carbons
X
H3C
CH3
H
H
H2C
most abundant E1 product
CH3
CH3
H3C
H
H
CH3
H2C
least abundant E1 product
fig 18
Notice in the figure above that the three possible E1 products do not form in equal
abundance. The most abundant alkene product is that which is most substituted: in other
words, the alkene in which the two sp2 carbons are bonded to the fewest hydrogen atoms.
In this case, the most substituted alkene has zero hydrogen substituents. The least
substituted - and least abundant - alkene product has two hydrogen substituents, while the
middle alkene has one hydrogen substituent. This trend is observed generally with
nonenzymatic E1 elimination reactions, and is known as Zaitsev's rule after the Russian
chemist Alexander Zaitsev.
226
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
14.2C: Stereochemistry of E1 elimination
Nonenzymatic E1 reactions can also result in both cis and trans alkenes. Keeping in mind
that in general trans alkenes are more stable than cis alkenes, we can predict that trans
alkenes will predominate in the product mixture.
X
+
minor E1 product
major E1 product
fig 19
Exercise 14. 8: Draw the structures of all possible E1 products starting with the
compounds below, and rank them in order of highest to lowest abundance.
tosylate group: see section 8.4
CH3
Cl
OTs
Br
fig 20
14.2D: The E2 elimination mechanism
When a strong base is combined with an alkyl halide (or alkyl tosylate/mesylate),
elimination generally occurs by the E2 pathway, in which proton abstraction and loss of
the leaving group occur simultaneously, without an intervening carbocation intermediate:
bond-breaking and bond-forming occurs
simultaneously, no carbocation intermediate
H
H C
H
H3C C
Br
H3C
CH3
O C
CH3
CH3
H
H3C
C
C
H
Br
CH3
C(CH3)3OH
strong base
fig 20a
Just like in the SN2 mechanism, the '2' in the E2 designation refers to the idea that the
rate-determining (and only) step of the reaction is a collision between the two reacting
molecules, in this case bromocyclohexane and methoxide ion.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
227
Chapter 14: Electrophilic reactions
14.2E: Competition between elimination and substitution
Consider a reaction between water and bromocyclohexane. Based on what we have just
learned, a likely product would be the alkene formed from an E1 elimination reaction
(pathway (a) in red below).
OH
water acts as nucleophile
OH2
b
substitution
Br
b
H
a
a
OH2
water acts as base
elimination
fig 20c
However, the reaction could take another course: what if the water molecule, instead of
acting as a base, were to act as a nucleophile (pathway (b) in blue? This should look
familiar - it is simply an SN1 reaction (section 8.1B). In fact, the reaction would result in
a mixture of elimination (E1) and substitution (SN1) products. This is a common theme:
elimination and substitution often compete with each other.
When both elimination and substitution products are possible, however, we can often
predict which reaction will predominate. In general, strong bases favor elimination, and
powerful nucleophiles favor substitution.
228
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
Br
I
+
I
weak base
good nucleophile
mainly substitution product
CH3
Br
+
O C CH3
CH3
mainly elimination product
strong base
poor nucleophile (hindered)
OH
Br
+
+
HO
strong base
good nucleophile
mixture of products
OH
Br
+
+
H2O
weak base
poor nucleophile
mixture of products
fig 20b
In addition, primary alkyl halides are much more likely to undergo substitution than
elimination, even when the nucleophile is also a strong base, because the electrophilic
carbon is unhindered and accessible to the nucleophile. Recall that the Williamson ether
synthesis (section 8.9A) is an efficient laboratory SN2 reaction between a primary (or
methyl) alkyl halide and an alkoxide. If a secondary alkyl halide is used, a substantial
amount of elimination product will form (the electrophilic carbon is more hindered, and
the alkoxide will act as a base as well as a nucleophile).
Br
+
OCH3
CH3O
substitution only
primary
OCH3
Br
+
+
CH3O
mixture of products
secondary
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
229
Chapter 14: Electrophilic reactions
fig 20e
While competition between substitution and elimination pathways is an issue for
chemists running reactions in the lab, the same cannot be said of biochemical reactions,
as the architecture active site of enzymes have evolved to ensure the formation of only
one product.
Exercise 14.9: Predict the major organic product(s) of the following reactions. If the
reaction is expected to result in a mixture of elimination and substitution product, show
both.
a) bromocyclopentane plus ethoxide
b) 1-chlorohexane plus CH3Sc) 2-iodo-2-methylpentane plus cyclopentanol
14.2F: Biochemical E1 elimination reactions
Looking through metabolic pathways in a biochemistry textbook, you'll see that almost
all elimination reactions appear to be of the E1cb type, occurring on carbons in the a,b
position relative to a carbonyl or imine. A relatively small number of elimination steps,
however, take place away from the electron-withdrawing influence of a carbonyl or
imine, and these are of the carbocation-intermediate, E1 type. The E2 mechanism is veyr
rare in biochemical pathways.
A reaction in the histidine biosynthetic pathway (EC 4.2.1.19) provides an example of a
biological E1 dehydration step:
PO
HO H
NH
N
resonance-stabilized
carbocation
H A
OH
PO
HO H
NH
N
PO
HO H
NH
PO
N H
OH
N
N
B:
tautomerization
PO
N H
O
fig 21a
230
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
N
Chapter 14: Electrophilic reactions
Notice that an E1cb mechanism is not possible here - there is no adjacent carbonyl or
imine and thus no possibility for a stabilized anionic intermediate. Instead, the first step
is loss of water to form a resonance-stabilized carbocation intermediate. Deprotonation
completes the E1 phase of the reaction to form an enol, which rapidly tautomerizes to a
ketone.
Another example of a biological E1 reaction is found in the biosynthesis pathway for
aromatic amino acids (EC 2.5.1.19):
CO2
CO2
Pi
OP
PO
O
OH
CO2
CO2
PO
O
OH
CH3
CH2
fig 22
Exercise 14.10: Draw a complete mechanism for the reaction above. Show how the
carbocation intermediate is stabilized by resonance.
Exercise 14.11 : Another step (EC 4.2.3.5) in the aromatic acid biosynthesis pathway
could be referred to as a conjugated E1 elimination of phosphate, the mechanistic reverse
of electrophilic addition to a conjugated diene (section 14.1E).
CO2
CO2
Pi
CO2
PO
CO2
O C
OH
O C
CH2
OH
CH2
chorismate
fig 23
a) Draw a complete mechanism for this reaction, showing two resonance contributors of
the carbocation intermediate.
b) An alternate E1 product can be envisioned for the starting compound above. Draw
this hypothetical E1 product, and explain why it is less stable than chorismate, the actual
product.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
231
Chapter 14: Electrophilic reactions
In section 13.3, we saw some Claisen condensation reactions in which the usual protonabstracton step was replaced by decarboxylation. A similar thing can happen with E1
eliminations:
decarboxylation instead of
deprotonation
(X = leaving group)
O
X
CO2
O
X
C O
R C C R
R
R R
R
R
C O
C C R
R
C C
R
R
R
fig 24
Isopentenyl diphosphate, the 'building block' for all isoprenoid compounds, is a product
of this type of hybrid decarboxylation / elimination reaction (EC 4.1.1.33).
Pi
CO2
OP
PPO
PPO
O
PPO
O
O
O
fig 25
Exercise 14.12: A conjugated decarboxylation/E1 elimination reaction (EC 4.2.1.51)
occurs in the phenylalanine biosynthesis pathway.
O
CO2
H2O
O
CO2
HO
O
fig 26
a) Predict the product.
b) Speculate as to why this reaction does not require the involvement of ATP or another
phosphate donor.
Section 14.3: Electrophilic alkene isomeration
Electrophilic reactions in biochemistry are not limited to addition to alkene double bonds.
The position of a double bond in an alkene can also be shifted through an electrophilic,
carbocation-intermediate reaction. An electrophilic alkene isomerization occurs when an
232
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
initial π bond protonation event (step 1 below) is followed by deprotonation of an
adjacent carbon to re-form the π bond in a different location.
Electrophilic isomerization mechanism
H H
R1
H A
B:
H
H H H H
position of double bond has shifted
R1
R2
H
R2
R1
H H
R2
fig 27
In a key early step in the biosynthesis of isoprenoid compounds, isopentenyl diphosphate
(IPP), the isoprenoid 'building block' molecule, is isomerized to dimethylallyl
diphosphate (DMAPP) (EC 5.3.3.2).
enz
S
H
H
PPO
PPO
H H H H
IPP
PPO
DMAPP
O
O
enz
fig 28
In the first step, the π bond between carbon #3 and carbon #4 is protonated by a cysteine
residue in the active site. X-ray crystallography studies on the isomerase enzyme (EMBO
J. 2001, 20, 1530) show that the carbocation intermediate is bound in a very deep,
hydrophobic active site cavity that seals out any water molecules that could potentially
attack the carbocation to form an undesired alcohol product. Instead, a basic glutamate
residue is positioned in the active site to abstract a proton from carbon #2 (step 2),
serving to reestablish the double bond in a new position between carbons #2 and #3.
Section 14.4: Electrophilic substitution
We have already been introduced to electrophilic addition and electrophilic isomerization
- now, let's move to the third variation on the electrophilic theme, that of electrophilic
substitution. In an electrophilic substitution reaction, a pair of π-bonded electrons first
attacks an electrophile - usually a carbocation species - and a proton is then abstracted
from an adjacent carbon to reestablish the double bond, either in the original position or
with isomerization.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
233
Chapter 14: Electrophilic reactions
Electrophilic substitution mechanism:
R3
R3
R1
R2
step 1
R1
R2
then . . .
:B
H
R1
R3
R3
R2
or . . .
step 2
(no isomerization)
R1
:B
H
R1
R2
R3
R3
R2
step 2
(isomerization)
R1
R2
fig 29
14.4A: Electrophilic substitution reactions in isoprenoid biosynthesis
Electrophilic substitution steps are very important in the biosynthetic pathways if
isoprenoid compounds. In an early, chain-elongating reaction (EC 2.5.1.1) of the
pathways of many isoprenoids, building blocks IPP and DMAPP combine to form a 10carbon isoprenoid product called geranyl diphosphate (GPP):
new C-C bond
PPi
+
PPO
IPP
PPO
PPO
DMAPP
GPP
fig 30
In a preliminary step (step a below), the diphosphate group on DMAPP departs to form
an allylic carbocation.
234
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
PPO
step a
PPO
PPO
step 1
H
allylic carbocation
:B
step 2
(with isomerization)
PPO
GPP
fig 32
In step 1, the π electrons in IPP then attack the electrophilic carbocation from step a,
resulting in a new carbon-carbon bond and a tertiary carbocation intermediate. Proton
abstraction (step 2) leads to re-establishment of a double bond one carbon over from
where it started out in IPP.
Exercise 14.13: IPP is much more stable than DMAPP when they are dissolved in water.
Explain why.
Exercise 14.14: Farnesyl diphosphate (FPP) is synthesized by adding another five-carbon
building block to geranyl diphosphate. What is this building block - IPP or DMAPP?
Draw a mechanism for the extension of GPP to FPP.
PPO
farnesyl diphosphate
fig 33a
Exercise 14.15: Propose a likely mechanism for the following transformation, which is
the first stage in a somewhat complex reaction in the synthesis of an isoprenoid
compound in plants. (Science 1997, 277, 1815)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
235
Chapter 14: Electrophilic reactions
PPi
OPP
farnesyl diphosphate
fig 33
Exercise 14.16: The electrophilic carbon in an electrophilic substitution reaction is often
a carbocation, but it can also be the methyl group on S-adenosylmethionine (SAM - see
section 8.8A). Propose a likely mechanism for this methylation reaction. (Biochemistry
2012, 51, 3003)
SAM
SAH
PPO
PPO
gernanyl diphosphate
2-methylgeranyl diphosphate
14.4B: Electrophilic aromatic substitution
Until now, we have been focusing mostly on electrophilic reactions of alkenes. Recall
from section 2.2C that π bonds in aromatic rings are substantially less reactive than those
in alkenes. Aromatic systems, however, do in fact undergo electrophilic substitution
reactions given a powerful electrophile such as a carbocation, and if the carbocation
intermediate that forms can be sufficiently stabilized.
Electrophilic aromatic substitution (Friedel-Crafts alkylation) mechanism
R
R
H
R
R
H
R
R
R
R
R
R
R
:B
R
R
R
R
R
R
R
fig 34
Organic chemists often refer to electrophilic aromatic substitution reactions with
carbocation electrophiles as Friedel-Crafts alkylation reactions.
Exercise 14.17: Aromatic rings generally do not undergo electrophilic addition reactions.
Why not?
236
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
The Friedel-Crafts reaction below is part of the biosynthesis of vitamin K and related
biomolecules.
O
OH
carbocation intermediate is stabilized by
resonance with oxygen lone pairs
H
OH
PPO
step a
O
OH
O
OH
O
OH
step 2
step 1
OH
OH
H
OH
B:
fig 36
Loss of diphosphate creates a powerful carbocation electrophile (step a) which attracts
the π electrons of the aromatic ring to form a carbocation intermediate with a new
carbon-carbon bond (step 1). Substitution is completed by proton abstraction (step 2)
which re-establishes the aromatic sextet.
An important point must be made here: because aromatic π bonds are substantially less
reactive than alkene π bonds, the electrophilic must be VERY electrophilic - usually a
carbocation. In addition, the carbocation intermediate that results from attack by
aromatic π electrons is generally stabilized by resonance with lone pair electrons on a
nearby oxygen or nitrogen (look at the resonance forms of the positively-charged
intermediate that forms as the result of step 1 in the above figure).
Remember that stabilizing the intermediate formed in a rate-limiting step has the effect of
lowering the activation energy for the step, and thus accelerating the reaction.
Organic chemists use the term ring activation to refer to the rate-accelerating effect of
electron-donating heteroatoms in electrophilic aromatic substitution reactions. Aromatic
rings lacking any activating oxygen or nitrogen atoms are less reactive towards
electrophilic substitution.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
237
Chapter 14: Electrophilic reactions
An example of the ring-activating effect of the nitrogen atom on an aromatic ring can be
found in the following Friedel-Crafts reaction (EC 2.5.1.34), which should be familiar
from the introduction to this chapter:
CO2
CO2
NH3
NH3
PPi
+
OPP
N
H
tryptophan
N
H
dimethylallyl tryptophan
DMAPP
fig 37
Recall that this is a key early step in the biosynthetic pathway for the ergot alkaloids
which are hypothesized to have been the root cause of the 'bewitchment' of several young
girls in 17th century Salem, Massachusetts. (J. Am. Chem. Soc. 1992,114, 7354).
Exercise 14.18: Draw a likely mechanism for the biosynthesis of dimethylallyl
tryptophan, including a resonance structure showing how the carbocation intermediate in
the rate determining step is stabilized by lone pair electrons on the ring nitrogen (in other
words, show how the nitrogen serves to activate the ring).
Friedel-Crafts reactions, in addition to being important biochemical transformations, are
commonly carried out in the laboratory. It is instructive to consider a few examples to
see how the same principles of structure and reactivity apply to both biochemical and
laboratory reactions.
Below is an example of a laboratory Friedel-Crafts alkylation reaction:
+
Cl
AlCl3
+ HCl
fig 37a
Recall that a powerful electrophile - such as a carbocation - is required for an
electrophilic aromatic substitution to occur. The 2-chloropropane reactant is
electrophilic, but not electrophilic enough to react with benzene. Here's where the
aluminum trichloride catalyst comes in: it reacts as a Lewis acid with the alkyl chloride to
generate a secondary carbocation:
238
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
Cl
Cl
Cl
Cl
Cl
Al
Al
Cl
Cl
Cl
fig 37b
The carbocation thus generated is sufficiently electrophilic to react with the aromatic π
electrons, in a manner that should be familiar from the biochemical examples discussed
above:
Cl
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
H
H
H Cl
fig 37c
You may have noticed, however, that one element from the biochemical Friedel-Crafts
reactions is missing here: there is no activating group to stabilize the ring carbocation
intermediate. Indeed, the presence of an activating group - for example, the oxygen atom
of a methoxy substituent - greatly increases the rate of a Friedel-Crafts alkylation.
ring-activating group
OCH3
OCH3
OCH3
+
Cl
+ HCl
+
AlCl3
ortho product
para product
fig 37d
Note in the example shown above that two products are formed: one is an orthodisubstituted benzene and one is para-disubstituted. Note also that no meta-disubstituted
product is formed. This phenomenon is referred to as the ortho-para directing effect,
and you are led towards an explanation in the exercise below.
Exercise 14.19:
a) Draw the lowest-energy resonance contributors of the carbocation intermediates
leading to formation of the ortho and para products in the reaction above. Your
structures should illustrate how the methoxy substituent is a ring-activating group.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
239
Chapter 14: Electrophilic reactions
b) Draw the hypothetical carbocation intermediate in a reaction leading to formation of a
meta-disubstituted product. Is this carbocation stabilized by the methoxy oxygen? Can
you see why no meta product forms?
Exercise 14.20:
a) Just as there are ring-activating groups in electrophilic aromatic substitutions, there are
also ring-deactivating groups. For each of the substituted benzene reactants below, draw
the carbocation intermediate and decide whether the substituent is ring-activating or ringdeactivating in a Friedel-Crafts reaction with 2-chloropropane and AlCl3 (in other words,
which compounds would react faster than benzene, and which would react slower?)
Explain how the ring-deactivating effect works.
O
H3C
NH
O
H
O
CH3
O
O
CH3
b) (challenging!) Ring-deactivating substituents are usually also meta-directing. Use one
of your carbocation intermediate drawings from part (a) of this exercise, and the concept
od resonance, to explain this observation.
c) (answer part (b) first) Look again at the vitamin K biosynthesis reaction in figure 36,
and discuss the ring activating/directing effects of the two substituents on the substrate.
Section 14.5: Carbocation rearrangements
Earlier in this chapter we introduced the so-called 'Markovnikov rule', which can be used
to predict the favored regiochemical outcome of electrophilic additions to asymmetric
alkenes. According to what we have learned, addition of HBr to 3-methyl-1-butene
should result in a secondary bromoalkane. However, the predominant product that is
actually be observed in this reaction is a tertiary alkyl bromide! Little or no secondary
alkyl bromide forms.
240
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
+ HBr
X
Br
Br
fig 38
To explain this result, let's take a look at the mechanism for the reaction:
Electrophilic addition with a hydride shift:
H Br
H 3C H
H 3C H
H
H 3C C
H 3C C
H
C C H
C C
step 1
H
H
H
H
:Br
H 3C
step 2
(hydride shift)
H H
C C C H
H 3C
H H
3
2
1
step 3
Br H H
H 3C
C
C C H
CH3 H H
Br
fig 39
Protonation of the double bond results in a secondary carbocation (step 1). What happens
next (step 2 above) is a process called a carbocation rearrangement, and more
specifically, a hydride shift. The electrons in the bond between carbon #3 and a
hydrogen are attracted by the positive charge on carbon #2, and they simply shift over to
fill the empty p orbital, pulling the proton over with them. Notice that the hydride, in
shifting, is not acting as an actual leaving group - a hydride ion is a very strong base and
a very poor leaving group.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
241
Chapter 14: Electrophilic reactions
An important reminder: a hydride ion (H-) is a proton plus two electrons. Be sure not to
confuse a hydride ion with H+, which is just a proton without any electrons.
As the shift proceeds, a new C-H σ bond is formed at carbon #2, and carbon #3 is left
with an empty p orbital and a positive charge.
CH3
H
H3C
H3C
H
H3C
H3C
H3C
H3C
CH3
H
H3C
H3C
H
H3C
H3C
CH3
H
H
CH3
H
H
CH3
3o carbocation
2o carbocation
fig 40
What is the thermodynamic driving force for this process? Notice that the hydride shift
results in the conversion of a secondary carbocation to a (more stable) tertiary
carbocation - a thermodynamically downhill step. As it turns out, the shift occurs so
quickly that it is accomplished before the bromide nucleophile has time to attack at
carbon #2. Rather, the bromide will attack (step 3) at carbon #3 to complete the addition.
Consider another example. When HBr is added to 3,3-dimethyl-1-butene, the product is
a tertiary - rather than a secondary - alkyl bromide.
3
+ HBr
3
2
X
Br
2
1
1
Br
3
2
1
fig 41
Notice that in the observed product, the carbon framework has been rearranged: the
methyl carbon indicated by a red dot has shifted from carbon #3 to carbon #2. This is an
example of another type of carbocation rearrangement, called a methyl shift.
Below is the mechanism for the reaction. Once again a secondary carbocation
intermediate is formed in step 1. In this case, there is no hydrogen on carbon #3 available
to shift over create a more stable tertiary carbocation. Instead, it is a methyl group that
242
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
does the shifting, as the electrons in the carbon-carbon σ bond move over to fill the
empty orbital on carbon #2 (step 2 below).
Electrophilic addition with methyl shift:
H 3C CH3
H Br
C
H 3C
3
C C
H
2
1
H 3C
H
H
H 3C
step 1
3
CH3
H
C C
1
2
H H H
H 3C
C
3
step 2
(methyl shift)
H
C C2 C1 H
3
H 3C
H H
3
3
2
2
1
:Br
CH3
1
2
1
step 3
H 3C
H 3C
Br CH3 H
C C2 C H
1
3
H H
Br
3
2
1
fig 42
The methyl shift results in the conversion of a secondary carbocation to a more stable
tertiary carbocation. The end result is a rearrangement of the carbon framework of the
molecule.
Exercise 14.21: In the (non-biochemcial) reactions below, the major product forms as the
result of a hydride or methyl shift from a carbocation intermediate. Predict the structure
of the major product for each reaction, disregarding stereochemistry.
a)
+
HBr
F
b)
+ H2O
H+
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
243
Chapter 14: Electrophilic reactions
c)
+
CH3OH
H+
d)
+
CH3OH
H+
(draw the substitution product)
Br
fig 43
Exercise 14.22: Draw the most abundant product of this laboratory Friedel-Crafts
reaction:
+
Cl
AlCl3
fig 43a
Carbocation rearrangements are involved in many known biochemical reactions.
Rearrangements are particularly important in carbocation-intermediate reactions in which
isoprenoid molecules cyclize to form complex multi-ring structures. For example, one of
the key steps in the biosynthesis of cholesterol is the electrophilic cyclization of
oxidosqualene to form a steroid called lanosterol (EC 5.4.99.7).
H
H
HO
O
(3S)-2,3-oxidosqualene
H
lanosterol
fig 44
This complex but fascinating reaction has two phases. The first phase is where the actual
cyclization takes place, with the formation of four new carbon-carbon bonds and a
carbocation intermediate. This phase is a 'cascade' of electrophilic alkene addition steps,
beginning with addition of an electrophilic functional group called an 'epoxide'.
The epoxide functional group - composed of a three membered ring with two carbons and
an oxygen - is relatively rare in biomolecules and biochemical reactions, and for this
reason it is not discussed in detail in this book. However, epoxides are an important and
244
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
versatile intermediate in laboratory organic synthesis, so you will learn much more about
how they are made and how they react if you take a course in chemical synthesis. For
now, it is sufficient to recognize that the carbon atoms of an epoxide are potent
electrophiles, due to both the carbon-oxygen bond dipoles and the inherent strain of the
three membered ring.
The second phase involves a series of hydride and methyl shifts culminating in a
deprotonation. In the exercise below, you will have the opportunity to work through the
entire cyclase reaction mechanism. In section 15.7, we will take a look at how the
epoxide group of oxidosqualene is formed. Trends Pharm. Sci. 2005, 26, 335; J. Phys
Chem B., 2012, 116, 13857.
Exercise 14.23:
a) The figure below outlines the first, cyclizing phase of the reaction that converts
oxidosqualene to lanosterol. However, the diagram is missing electron movement
arrows, and intermediates 1-4 are all missing formal charges - fill these in.
First phase (ring formation):
step 2
step 1
HO
O
(3S)-2,3-oxidosqualene
HO
intermediate 1
(missing formal charge)
intermediate 2
(missing formal charge)
step 3
continued in next
figure
step 4
HO
HO
intermediate 4
(missing formal charge)
intermediate 3
(missing formal charge)
(fig 20)
fig 45
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
245
Chapter 14: Electrophilic reactions
b) Next comes the 'shifting' phase of the reaction. Once again, supply the missing
mechanistic arrows.
Second phase: rearrangement and deprotonation
step 5
step 6
HO
HO
intermediate 4
(missing formal charge)
HO
intermediate 5
intermediate 6
step 7
H
H
step 8
step 9
HO
HO
H
lanosterol
HO
intermediate 8
intermediate 7
fig 46
c) Look at the first and last steps of the entire process: overall, would you describe this as
an electrophilic addition or substitution?
The oxidosqualene cyclization reaction and others like it are truly remarkable examples
of the exquisite control exerted by enzymes over the course of a chemical reaction.
Consider: an open-chain starting molecule is converted, by a single enzyme, into a
complex multiple fused-ring structure with seven chiral centers. Oxidosqualene could
potentially cyclize in many different ways, resulting in a great variety of different
products. In order for the enzyme to catalyze the formation of a single product with the
correct connectivity and stereochemistry, the enzyme must be able to maintain precise
control of the conformation of the starting compound and all reactive intermediates in the
active site, while also excluding water molecules which could attack at any of the
positively charged carbons.
246
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
Key learning objectives for this chapter
Understand why the π bond in a carbon-carbon double bond is more reactive than the σ
bond.
Addition
Be able to draw a mechanism for the electrophilic addition of a haloacid to an alkene.
Stereochemistry: understand why nonenzymatic electrophilic addition of a
haloacid to an alkene occurs with racemization (both inversion and retention of
configuration) at both alkene carbons. Be able to distinguish syn vs anti addition.
Regiochemistry: Be able to predict the regiochemical outcome of an electrophilic
addition, based on the relative stability of the two possible carbocation
intermediates. Be able to predict when anti-Markovnikov addition is likely to
occur.
Be able to predict the product of nonenzymatic addition of water/alcohol to an alkene,
including regio- and stereo-chemistry when applicable. Be able to draw complete
mechanisms.
Be able to predict the products of nonenzymatic addition of water/alcohol to a
conjugated diene or triene, including regio- and stereochemistry when applicable. Be
able to draw complete mechanisms, including multiple resonance forms for carbocation
intermediates.
Be able to apply your understanding of nonenzymatic alkene addition reactions to draw
mechanisms for enzymatic addition reactions. In particular, you should be able to draw
mechanisms for biochemical electrophilic addition reactions in which a new carboncarbon bond is formed.
Elimination
Be able to draw a mechanism for an E1 elimination reaction.
Be able to predict possible E1 reaction products from a common starting compound ,
taking into account both regiochemistry (Zaitsev's rule) and stereochemistry.
Be able to recognize and draw a mechanism for biochemical E1 reactions in which
a) the second step is a deprotonation event
b) the second step is a decarboxylation event
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
247
Chapter 14: Electrophilic reactions
Be able to distinguish whether a biochemical elimination reaction is likely to proceed
through a E1cb or E1 mechanism, based on the structure of the starting compound.
Isomerization/substitution
Be able to recognize and draw mechanisms for a biochemical electrophilic isomerization
reaction (shifting the location of the carbon-carbon double bond).
Be able to recognize and draw mechanisms for a biochemical electrophilic substitution
reaction.
Be able to recognize and draw mechanisms for a biochemical electrophilic aromatic
substitution reaction, and be able to explain the ring-activating effect (how the
carbocation intermediate is stabilized by resonance, usually with lone-pair electrons on
either an oxygen or a nitrogen atom).
Be able to recognize when a hydride or alkyl shift is likely to occur with a carbocation
reaction intermediate.
Be able to draw a mechanism for a reaction that includes a carbocation rearrangement.
248
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
Problems
P14.1: Draw the major product(s) (including all stereoisomers) that would be expected to
result from the nonenzymatic electrophilic addition reactions below. Your product(s)
should result from the most stable possible carbocation intermediate. Hint: consider the
possibility of thermodynamically favorable rearrangement steps.
a)
+
HBr
b)
+
HBr
D
c)
H+
+
H2O
+
CH3OH
d)
H+
e)
+
CH3CH2OH
H+
O
P14.2: Draw likely mechanisms for the nonenzymatic reactions below. Products shown
are not necessarily the most abundant for the reaction.
a)
Br
+
HBr
b)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
249
Chapter 14: Electrophilic reactions
Br
HBr
+
c)
Br
HBr
+
d)
OH
H+
P14.3: Provide mechanisms for the following reactions, both of which are part of an
alkaloid synthesis pathway in fungi. (Microbiol. 2005, 151, 2199)
O
N
N
H
O
HN
O
H
N
HN
N
H
O
H
O
+
PPO
N
N
H
HN
O
H
P14.4: Draw a likely mechanism for the reaction below. The product is myrcene, a
compound produced by fir trees as a defense against insects. (J. Biol. Chem 1997, 272,
21784)
250
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
PPO
PPi
P14.5: Provide a mechanism for the following reaction from the vitamin B12 biosynthetic
pathway, and identify the missing participants indicated by questions marks in the figure.
R
R
R
R
H
H3C R
R
R
N
N
N
N
R
R
?
?
R
R
H
R
H
H
P14.6: A diene molecule synthesized in the laboratory was found to irreversibly inhibit
the action of isopentenyl diphosphate isomerase (section x.x) when the carbon indicated
with a dot becomes covalently bonded to a cysteine residue in the enzyme's active site.
Propose a mechanism showing how this could happen. ( J. Am. Chem. Soc. 2005, 127,
17433)
PPO
inhibitor
P14.7: Nonenzymatic electrophilic addition of water to alkynes results in the formation
of a ketone or an aldehyde, depending on the starting alkyne. A vinylic carbocation is a
key intermediate, and the reaction is accelerated with the use of a catalytic amount of
strong acid. Predict the product the addition of water to ethyne, and draw a mechanism
for the reaction.
P14.8: The reaction below is part the pathway by which some bacteria -including the
species which cause tuberculosis and leprosy - form distinctive branched-chain fatty
acids for incorporation into their cell walls. This enzyme is of interest to scientists as
possible targets for new antibiotic drugs. Propose a likely mechanism, and identify the
missing participants denoted by questions marks. (J. Biol. Chem. 2006, 281, 4434)
O
?
O
?
HO
R
n
O
R
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
n
O
251
Chapter 14: Electrophilic reactions
P14.9: Suggest a likely mechanism for this reaction, which is a key step in the synthesis
of bacterial cell walls. Your mechanism should show an electrophilic addition, followed
by an E1 elimination.
OH
O2C
PO
CH2 +
OH
Pi
O
HO
HO
HN
O
O
H2C
O
CH3
O
HO
CO2
ribose-U
HN
O
O
CH3
ribose-U
P14.10: Suggest a mechanism for the following reaction, which is part of the pathway by
which many microbes synthesize methanopterin, a derivative of the vitamin folic acid.
Hint: the mechanism can be described as an electrophilic aromatic substitution with a
final decarboxylation step in place of the usual deprotonation step. (J. Biol. Chem. 2004,
279, 39389.
OP
O
H2N
CO2
OPP
PPi
CO2
NH2
OP
O
+
HO
OH
HO
OH
(
P14.11: Researchers investigated the mechanism of the enzyme 3-deoxy-D-mannooctulosonate-8-phosphate synthase by running the reaction with one of the substrates
labeled with the 18O isotope (colored red in the scheme below). Consider the two
hypothetical results shown below, each pointing to a different mechanism. Both
mechanisms involve a carbocation intermediate. (Biochem. Biophys. Res. Commun.
1988, 157, 816)
252
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
O
O P O
H2O
OH
O2C
O
mechanism A
O P O
18O
OH
O
OH
18O
OH
OP
+
OP
OH
O
O P O
O2C
O
OH
OH
H2O
OH
18O
OH
O2C
mechanism B
OP
O
OH
OH
a) Propose a mechanism that is consistent with result A
b) Propose a mechanism that is consistent with result B.
P14.12: Consider the following isomerization reaction (J. Biol. Chem. 1989, 264, 2075):
OPP
PPO
linalyl diphosphate
geranyl diphosphate
(JBC264, 2075)
a) Suggest a likely mechanism involving a carbocation intermediate.
b) Suggest an isotopic labeling experiment (using substrate labeled with 18O) that could
confirm or rule out a alternative, concerted isomerization mechanism (ie. one without
formation of a carbocation intermediate). Explain your reasoning.
c) Propose a mechanism for the following reaction (notice that the starting compound is
linalyl diphosphate from part (a), drawn in a different conformation). (Arch Biochem
Biophys 2003, 417, 203)
OPP
H2O
PPi
OH
linalyl diphosphate
O
1,8-cineole
(
For parts d-f, refer to the figure below:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
253
Chapter 14: Electrophilic reactions
OPP
PPi
PPi
OPP
(+)-bornyl diphosphate
linalyl diphosphate
(+)-sabinene
d) Provide mechanisms for the conversion of linalyl diphosphate to (+)-bornyl
diphosphate
e) Provide mechanisms for the conversion of linalyl diphosphate to (+)-bornyl
diphosphate (+)-sabinene.
f) Is the second step in the (+)-bornyl diphosphate pathway (addition of phosphate) a
Markovnikov or anti-Markovnikov addition? Explain the regiochemistry of this step in
terms of carbocation stability.
P14.13: The two compounds shown below were each treated with HBr, and the products
isolated and analyzed by 1H NMR. Use the NMR data provided to determine the
structure of both products, then explain the observed regiochemistry of the addition
reaction.
O
methyl vinyl ketone
O
OCH3
methyl methacrylate
1
H-NMR data for product of HBr addition to methyl vinyl ketone:
d (ppm)
2.2
3.0
3.5
254
integration
1.5
1
1
splitting
s
t
t
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
1
H-NMR data for product of HBr addition to methyl methacrylate:
d (ppm)
integration
splitting
1.3
2.3
3.5
3.7
3
1
2
3
d
sextet
d
s
P14.14: Ketones and aldehydes with a hydroxy group in the a position are known to
undergo an isomerization reaction known as an acyloin rearrangement:
A
H
O
HO
R
R
O
R
O
R
H O R
=
R
R
R
R
OH
:B
Notice in the general acyloin rearragnement mechanism below, the green alkyl group is
shifting from the a carbon (red) to the carbonyl carbon (blue). Notice also that this shift
does not involve a carbocation intermediate, although a resonance contributor can be
drawn in which the carbonyl carbon has a positive charge.
a) Draw a mechanism for this acyloin rearrangement step in the biosynthetic pathway for
the amino acid leucine:
O
O
O2C
H3C
H3C OH
HO CO2
b) Draw a mechanism for this acyloin rearrangement step in the isoprenoid biosynthetic
pathway in bacteria:
O
OH
O
OP
H3C
HO H
OH
OP
H
H3C OH
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
255
Chapter 14: Electrophilic reactions
P14.15: Propose mechanisms for these reactions in the vitamin B12 biosynthetic pathway:
a)
O2C
O2C
CO2
CO2
H2O
O2C
O2C
+
R1
N
N
H
H
CO2
CO2
R
R2
HO
N
N
H
H
R
b)
R
R
CH3
N
N
CH3
CH3
CH3
R
R
R
R
c)
R
R
R
R
R
NH4+
R
R
R
+
R
N
H
N
H
R
NH3 H
H
N
N
H
H
H
)
P14.16: An early reaction in the biosynthesis of tryptophan can be described as an
intramolecular electrophilic aromatic substitution/decarboxylation hybrid, followed by an
E1 dehydration (EC 4.1.1.48).
O
CO2
H2O
O
OH
N
O
OP
OP
N
H
OH
OH
OH
H
a) Draw a mechanism that corresponds to the verbal description given above. Use
resonance structures to show how the nitrogen atom helps to stabilize the carbocation
intermediate. Hint: the electrophilic carbon in this case is a ketone rather than a
carbocation.
256
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
b) What aspect of this reaction do you think helps to compensate for the energetic
disadvantage of not having a powerful carbocation electrophile?
c) Again thinking in terms of energetics, what is the 'driving force' for the dehydration
step?
P14.17: Propose a likely carbocation-intermediate mechanism for the following reaction
in the biosynthesis of morphine, being sure to identify the structure of the organic
compound released in the reaction.(
H3C
O
H3C
?
O
O
O
N
N
CH3
CH3
O
O
OH
P14.18: Propose mechanisms for these three electrophilic cyclization reactions.
Carbocation rearrangement steps are involved.
a) epi-arisolochene
b) vetispiradiene (Science 1997, 277, 1815)
PPi
epi-arisolochene
OPP
farnesyl diphosphate
common intermediate
vetispiradiene
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
257
Chapter 14: Electrophilic reactions
c) pentalenene (Science 1997, 277, 1820)
PPi
PPO
pentalenene
intermediate
farnesyl diphosphate
P14.19: Strictosinide, an intermediate in the biosynthesis of the deadly poison strychnine,
is formed from two steps: a) intermolecular imine formation, and b) an intramolecular,
ring-forming electrophilic aromatic substitution with the imine carbon from step (a) as
the electrophile.
O
glucose
NH
O
N
H
O
OCH3
strictosinide
Given this information, predict the two precursors to strictosinide, and draw a mechanism
for the reaction described. Hint: use the 'retro' skills you developed in chapters 12 and 13.
P14.20: In the introduction to chapter 8, we learned about reactions in which the
cytosine and adenine bases in DNA are methylated. In the course of that chapter, we
learned how adenine N-methylation occurs in bacteria, but we were not yet equipped to
understand cytosine C-methylation, which was the more relevant reaction in terms of
human health and development. Now we are: propose a reasonable mechanism for the Cmethylation of cytosine as shown in figure 8.x.
P14.21: The reaction below has been proposed to proceed via a cyclization step followed
by an E1/decarboxylation step (in other words, an E1 mechanism where decarboxylation
occurs instead of deprotonation). Draw a mechanism that fits this description, and show
the most stable resonance contributors of the two key cationic intermediates.
(J. Am. Chem. Soc. 2009, 131, 14648 (Scheme 2 pathway A)
258
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 14: Electrophilic reactions
O
OH
H2O
CO2
O
N
H
HO CO2
N
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OH
259
Chapter 15: Oxidation and reduction
Chapter 15
Oxidation and reduction reactions
Introduction
Theo Ross was not doing very well at his summer job, and he was frustrated. His boss
had given him specific instructions, and yet Theo kept botching the job, over and over
again. Theo was not used to failure – he had achieved almost perfect scores on both the
ACT and SAT college entrance exams, and was headed to Stanford University in the fall.
Why couldn't he get it right? It wasn't brain surgery, after all.
Well, actually – it was brain surgery.
An April 17, 2014 article in Sports Illustrated tells Theo's story. Wanting to do
something interesting over the summer of 2010 before he started college, Theo had
applied for a research internship at the National Institutes of Health in Bethesda,
Maryland. This was an extremely competitive program normally reserved for outstanding
college students, but somehow Theo had managed to win a coveted spot in the program,
working with Dr. Dorian McGavern, a neurologist studying how meningitis effects the
brain. Dr. McGavern assigned Theo the task of performing 'skull-thinning' surgery on
mice, part of which involved using a special saw to shave down a small section of the
bone in order to gain access to the brain. It was a delicate procedure, something that even
some experienced neurosurgeons who had tried it had found challenging. Any small slip
resulted in a concussion to the mouse's brain, rendering it useless for the study. Theo just
couldn't get the hang of it, and ended up concussing one mouse after another.
You have probably heard the old expression: “when life gives you lemons, make
lemonade”. Theo made a lot of lemonade that summer.
260
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Theo and Dr. McGavern eventually realized that his failure at the procedure actually
presented an opportunity to observe what happens to a brain right after a concussive
injury. Theo started doing more skull-thinning surgeries, but now the goal was to cause
concussion, rather than to avoid it. The concussed mice (who had been anaesthetized
prior to the surgery) were immediately strapped under a microscope so that Theo could
observe how their brains responded to the injury. This was new, and very exciting stuff:
most of what neurologists knew about concussions up to that point had come from MRI
(magnetic resonance imaging – see chapter 5) or autopsies. Nobody knew very much
about what happens at the cellular level in a brain in the minutes and hours after a
concussion has occurred. In addition, the problem of traumatic head injuries and the longlasting effects they cause was becoming an increasingly hot topic in the news, critically
relevant to thousands of veterans returning from Iraq and Afghanistan as well as to
football players and other athletes in contact sports– including Theo, who had been a
competitive wrestler in high school. (You might have been wondering why Theo's story
appeared in Sports Illustrated – now you know.)
Theo spent the rest of that summer, and every spring break and summer vacation over the
next few years, working in McGavern's lab on the new project. He and McGavern found
evidence that the 'hidden' damage to a concussed brain – that which went undetected in
MRI scans but could come back to haunt the victim years later in the form of recurring
headaches, memory loss, and depression – may be caused by a type of molecule referred
to as 'reactive oxygen species', or ROS, leaking from damaged tissues into the brain. ROS
are potentially harmful byproducts of respiration such as hydrogen peroxide (H2O2) that
are constantly being produced in our cells. Although ROS can cause serious oxidative
damage if they are allowed to build up, our bodies have evolved ways to deal with them,
using so-called 'ROS scavengers' to convert them to something innocuous like water.
With this new understanding, Theo and his mentor had another idea: what if they could
prevent the ROS from causing further damage to a recently concussed brain by applying
an scavenger to the injury? After some trial and error, they found that an ROS scavenger
compound called glutathione, when applied directly to the skull of a concussed mouse
within a few minutes to three hours after the injury, could permeate the bone and react
with the ROS. Brain cells from these glutathione-treated mice appeared normal, with
none of the signs of ROS damage Theo was used to seeing.
NH3
H
O
N
O2C
N
CO2
H
O
HS
glutathione
fig 1a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
261
Chapter 15: Oxidation and reduction
The road from an initial scientific discovery to a safe and effective medical treatment is
often a very long one, but Theo Roth and Dorian McGavern appear to have made a
discovery that could eventually help prevent some of the most devastating and long-term
damage caused by traumatic head injuries. In the end, it's a very good thing that Theo's
hands were not cut out for brain surgery.
***
The chemistry of oxidation and reduction - often called 'redox' chemistry - is central to
Theo Roth's discovery about what happens to a concussed brain at the molecular level.
This chapter is dedicated to redox chemistry. We'll begin with a reminder of what you
learned in General Chemistry about the fundamentals of redox reactions in the context of
inorganic elements such as iron, copper and zinc: reduction is a gain of electrons, and
oxidation is a loss of electrons. Then, we'll expand our understanding to include
bioorganic redox reactivity, examining among other things how alcohols are converted to
ketones and aldehydes, aldehydes are converted to carboxylic acids, and amines are
converted to imines. We will also talk about redox reactions in the broader context of
metabolism in living things.
A central player in some of the biochemical redox reactions we will see is the coenzyme
called glutathione, Theo Roth's 'magic bullet' molecule that was able to rescue mouse
brain cells from death by oxidation. We'll see how glutathione acts as a mediator in the
formation and cleavage of disulfide bonds in proteins, and how it acts as an 'ROS
scavenger' to turn hydrogen peroxide into water.
Section 15.1: Oxidation and reduction of organic compounds - an overview
You are undoubtedly already familiar with the general idea of oxidation and reduction:
you learned in general chemistry that when a compound or element is oxidized it loses
electrons, and when it is reduced it gains electrons. You also know that oxidation and
reduction reactions occur in tandem: if one species is oxidized, another must be reduced
at the same time - thus the term 'redox reaction'.
Most of the redox reactions you have seen previously in general chemistry probably
involved the flow of electrons from one metal to another, such as the reaction between
copper ion in solution and metallic zinc:
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq)
Exercise 15.1: Reading the reaction above from left to right, which chemical species is
being oxidized? Which is being reduced?
When we talk about the oxidation and reduction of organic compounds, what we are
mainly concerned with is the number of carbon-heteroatom bonds in the compound
262
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
compared to the number of carbon-hydrogen bonds. (Remember that the term
'heteroatom' in organic chemistry generally refers to oxygen, nitrogen, sulfur, or a
halogen).
Oxidation of an organic compound results an increase in the number of carbonheteroatom bonds, and/or a decrease in the number of carbon-hydrogen bonds.
Reduction of an organic compound results in a decrease in the number of carbonheteroatom bonds, and/or an increase in the number of carbon-hydrogen bonds.
Below are a number of common functional group transformations that are classified as
redox.
OH
R C
H
H
O
oxidation
R
reduction
C
O
oxidation
H
R
reduction
aldehyde
alcohol
NH2
R C
NH
R
reduction
H
H H
H H
alkane
C
H
imine
amine
R C C
OH
carboxylic acid
oxidation
H
C
H
oxidation
reduction
R
H
C C
H
H
alkene
fig1
Heteroatoms such as oxygen and nitrogen are more electronegative than carbon, so when
a carbon atom gains a bond to a heteroatom, it loses electron density and is thus being
oxidized. Conversely, hydrogen is less electronegative than carbon, so when a carbon
gains a bond to a hydrogen, it is gaining electron density, and thus being reduced.
Exercise 15.2: The hydration of an alkene to an alcohol is not classified as a redox
reaction. Explain.
For the most part, when talking about redox reactions in organic chemistry we are dealing
with a small set of very recognizable functional group transformations. The concept of
oxidation state can be useful in this context. When a compound has lots of carbonhydrogen bonds, it is said to be in a lower oxidation state, or a more reduced state.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
263
Chapter 15: Oxidation and reduction
Conversely, if it contains a lot of carbon-heteroatom bonds, it is said to be in a higher
oxidation state.
We'll start with a series of single carbon compounds as an example. Methane, in which
the carbon has four bonds to hydrogen, is the most reduced member of the group. The
compounds become increasingly oxidized as we move from left to right, with each step
gaining a bond to oxygen and losing a bond to hydrogen. Carbon dioxide, in which all
four bonds on the carbon are to oxygen, is in the highest oxidation state.
OH
H
H C
H
H C
O
H
H
H
alkane
alcohol
H
C
O
H
aldehyde
H
C
OH
carboxylic acid
most reduced
O C O
carbon dioxide
most oxidized
fig 3
More generally, we can rank the oxidation state of common functional groups:
alkane
alcohol
thiol
amine
alkene
aldehyde
ketone
imine
carboxylic acid derivative
most reduced
most oxidized
fig 4
The alkane oxidation state is the most reduced. Alcohols, thiols, amines, and alkenes are
all at the same oxidation state: therefore, a reaction converting one of these groups to
another - an alcohol to alkene conversion, for example - is not a redox reaction.
Aldehydes, however, are at a higher oxidation state than alcohols, so an alcohol to
aldehyde conversion is an oxidation. Likewise, an imine to amine conversion is a
reduction, but an imine to ketone conversion is not a redox reaction.
It is important to keep in mind that oxidation and reduction always occurs in tandem:
when one compound is oxidized, another compound must be reduced. Often, organic
chemists will use the terms oxidizing agent and reducing agent to refer to species that
are commonly used, by human chemists or by nature, to achieve the oxidation or
reduction of a variety of compounds. For example, chromium trioxide (CrO3) is a
laboratory oxidizing agent used by organic chemists to oxidize a secondary alcohol to a
ketone, in the process being reduced to H2CrO3. Sodium borohydride (NaBH4) is a
laboratory reducing agent used to reduce ketones (or aldehydes) to alcohols, in the
process being oxidized to NaBH3OH.
264
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
alcohol gets oxidized in the reaction
oxidizing agent
(gets reduced in the reaction)
OH
R C
O
+
H
O
R
O
R
H
O
C
Cr
H B
R
OH
O
C
R
reducing agent
(gets oxidized in the reaction)
ketone gets reduced in the reaction
R
Cr lost a bond to oxygen
Na
O
Cr
boron lost a bond to hydrogen,
gained a bond to oxygen
H
OH
H
R C
H
OH
H B OH
H
R
Na
H
There is a wide selection of oxidizing and reducing agents available for use in the organic
chemistry laboratory, each with its own particular properties and uses. For example,
while sodium borohydride is very useful for reducing aldehyde and ketone groups to
alcohols, it will not reduce esters and other carboxylic acid derivatives. If you take a
course in synthetic organic chemistry, you will learn about the use of many of these
agents.
In this book, of course, we are concerned primarily with the organic chemistry that occurs
within a living cell. A large part of this chapter will be spent looking at the action of two
very important classes of coenzymes - the nicotinamides and the flavins - that serve as
biochemical oxidizing and reducing agents. We also consider the oxidation and reduction
of sulfur atoms in thiol groups, especially the thiol group on the side chain of cysteine
residues in proteins.
Exercise 15.3: Each of the biochemical transformations shown below is a step in amino
acid metabolism. For each, state whether the substrate is being oxidized, reduced, or
neither oxidized nor reduced.
a) (from aromatic amino acid biosynthesis)
CO2
CO2
O
OH
HO
OH
OH
OH
5-dehydroskikimate
shikimate
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
265
Chapter 15: Oxidation and reduction
b) (from the biosynthesis of arginine and proline)
NH3
NH3
O
O
CO2
+
CO2
Pi
H
OP
c) (from the catabolism of lysine)
O
O
O2C
SCoA
O2C
SCoA
d) (from the catabolism of tryptophan)
NH3
O
NH3
O
CO2
O
CO2
e) (from the catabolism of serine)
O
NH3
HO
O
O
serine
O
H3C
O
pyruvate
Section 15.2: Oxidation and reduction in the context of metabolism
Think back again to the redox chemistry that you learned in your general chemistry
course. A common experiment in a general chemistry lab is to set up a galvanic cell
consisting of a copper electrode immersed in an aqueous copper nitrate solution,
connected by a wire to a zinc electrode immersed in an aqueous zinc nitrate solution.
266
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
(http://chemwiki.ucdavis.edu)
When the cell is completed with a salt bridge, an electrical current begins to flow - what
we have is a simple battery (figure a above). Over time, the copper electrode gets heavier
as metallic copper is deposited on the copper cathode, while the zinc anode slowly
dissolves into solution (figure b above). The redox reaction occurring here is:
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq) + energy
Electrons flow from zinc metal to copper cations, creating zinc cations and copper metal:
in other words, zinc metal is being oxidized to zinc cation and copper cation is being
reduced to copper metal, as expressed by the two relevant half-cell reactions:
Cu+2(aq) + 2e- → Cu0(s)
Zn0(s) → Zn+2(aq) + 2eWe can predict before we set up the cell that the spontaneous flow of electrons will go in
the zinc to copper direction, just by looking at a table of standard reduction potentials
(such a table was no doubt in your general chemistry text).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
267
Chapter 15: Oxidation and reduction
Standard reduction potentials at 25 oC
Reduction half-reaction
Ag+1(aq) + e- → Ag0(s)
Reduction potential (volts)
0.800
Cu+2(aq) + 2e- → Cu0(s)
0.337
H+1(aq) + 2e- → H2(g)
Pb+2(aq) + 2e- → Pb0(s)
0
(standard)
-0.126
Fe+2(aq) + 2e- → Fe0(s)
-0.441
Zn+2(aq) + 2e- → Zn0(s)
-0.763
Copper ion (Cu+2) has a higher standard reduction potential than zinc ion (Zn+2), meaning
that, under identical conditions, more energy is released by reducing one mole of Cu+2
ion to Cu0 metal than is released by reducing one mole of Zn+2 ion to Zn0 metal.. Another
way to think about this is to imagine that the copper ion 'wants' to gain electrons more
than the zinc ion does. Conversely, zinc metal 'wants' to lose electrons more than the
copper metal does. Therefore, transfer of two electrons from zinc metal to Cu2+ is a
thermodynamically downhill process, whereas the reverse process - transfer of two
electrons from copper metal to Zn2+ - is thermodynamically uphill.
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq) + energy
Cu(s) + Zn+2(aq) + energy → Cu+2(aq) + Zn(s)
Let's now extend the idea of redox reactions to the context of metabolism in living things.
When we 'burn' glucose for energy, we transfer (by a series of enzyme-catalyzed
reactions) electrons from glucose to molecular oxygen (O2), oxidizing the six carbon
molecules in glucose to carbon dioxide and at the same time reducing the oxygen atoms
in O2 to water. The overall chemical equation is:
268
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
OH
O2 has been reduced to water
OH
O
+
HO
OH
6
O
O
6C
O
OH
H
+
+ energy
6O
O
H
all 6 carbons in carbon dioxide
oxidation state
5 carbons in alcohol oxidation state, one
carbon in aldehyde oxidation state
fig 8
The transfer of electrons from glucose to O2 is a thermodynamically downhill, energyreleasing process, just like the transfer of electrons from zinc metal to copper ion. And
while you could have used the energy released by the zinc/copper redox reaction to light
a small light bulb, your cells use the energy released by the glucose/oxygen redox process
to carry out a wide variety of energy-requiring activities, such as walking to your organic
chemistry lecture.
In your general chemistry copper/zinc experiment, was it possible to reverse the reaction
so that it runs in the uphill direction - in other words, to oxidize copper and reduce zinc?
Zn+2(aq) + Cu(s) + energy → Zn(s) + Cu+2(aq)
Just ask yourself the question: is it possible to get water to flow uphill? Of course it is but only if you supply a pump and some energy!
The same idea applies to 'pumping' electrons uphill in your copper-zinc electrochemical
cell: all you need to do is to provide some energy in the form of an external electrical
current in order to pump the electron flow in the uphill direction. You are recharging your
battery.
Thinking again in a biochemical context: plants are able, by a process called
photosynthesis, to reduce carbon dioxide and oxidize water to form glucose and
molecular oxygen: essentially recharging the ecosystem's biochemical battery using
energy from the sun.
6CO2 + 6H2O + energy → C6H12O6 + 6O2
On a global scale, oxidation of the carbons in glucose to CO2 by non-photosynthetic
organisms (like people) and the subsequent reductive synthesis of glucose from CO2 by
plants is what ecologists refer to as the 'carbon cycle'.
In general the more reduced an organic molecule is, the more energy is released when it
is oxidized to CO2 . Going back to our single-carbon examples, we see that methane, the
most reduced compound, releases the most energy when oxidized to carbon dioxide,
while formic acid releases the least:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
269
Chapter 15: Oxidation and reduction
energy of oxidation to CO2:
H
H C
H
methane
H
OH methanol
H C
H
H
O
H
-196 kcal/mol
C
-168 kcal/mol
-125 kcal/mol
formaldehyde
H
O formic acid
C
H
OH
-68 kcal/mol
O carbon dioxide
C
O
fig 9
A lipid (fat) molecule, where most of the carbons are in the highly reduced alkane state,
contains more energy per gram than glucose, where five of the six carbons are in the
more oxidized alcohol state (look again at the glucose structure we saw just a couple of
pages back).
O
O
O
most carbons are in the highly reduced alkane state
O
O
O
a saturated fat molecule
fig 10
After we break down and oxidize sugar and fat molecules to obtain energy, we use that
energy to build large, complex molecules (like cholesterol, or DNA) out of small, simple
precursors. Many biosynthetic pathways are reductive: the carbons in the large
biomolecule products are in a reduced state compared to the small precursors. Look at
the structure of cholesterol compared to that of acetate, the precursor molecule from
which all of its carbon atoms are derived - you can see that cholesterol is overall a more
reduced molecule.
Organic Chemistry With a Biological Emphasis (2016 ed.)
270
Tim Soderberg
Chapter 15: Oxidation and reduction
reduced final product
oxidized precursor molecule
O
H3C
C
O
acetate
HO
cholesterol
fig 11
While we are focusing here on the mechanistic details of the individual organic redox
reactions involved in metabolism, if you take a course in biochemistry you will learn
much more about the bigger picture of how all of these reactions fit together in living
systems.
Section 15.3: Hydrogenation of carbonyl and imine groups
Next, we'll go on to look at the actual chemical mechanisms involved in the enzymecatalyzed oxidation and reduction of biomolecules.
15.3A: An overview of hydrogenation and dehydrogenation
Many of the redox reactions that you will encounter when studying the central metabolic
pathways are classified as hydrogenation or dehydrogenation reactions. Hydrogenation
is simply the net addition of a hydrogen (H2) molecule to a compound, in the form of a
hydride ion (H-, a proton plus a pair of electrons) and a proton. Hydrogenation
corresponds to reduction. Dehydrogenation reactions are the reverse process: loss of a
hydride and a proton. Dehydrogenation corresponds to oxidation.
proton
hydrogenation
(reduction)
O
H3C C
H
O
H3C C
dehydrogenation
(oxidation)
H
H
H
hydride ion
(proton plus two electrons in C-H bond)
fig 2
Hydrogenation and dehydrogenation reactions can also be called hydride transfer
reactions, because a hydride ion is transferred from the molecule being oxidized to the
one being reduced. In the next few sections, we will learn about two important classes of
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
271
Chapter 15: Oxidation and reduction
coenzyme molecules that serve hydride ion acceptors (oxidizing agents) and hydride
ion donors (reducing agents) in biochemical redox reactions.
Be careful not to confuse the terms hydrogenation and dehydrogenation with hydration
and dehydration - the latter terms refer to the gain or loss of water, while the former
terms refer to the gain or loss of hydrogen.
Many mechanistic patterns that we have already learned about in previous chapters will
come into play again in this discussion, with the only variation being that here, a hydride
ion will act as a nucleophile (in the hydrogenation direction) or as a leaving group (in the
dehydrogenation direction). The key to understanding these reactions will be to
understand how a hydride can act as a nucleophile or leaving group.
15.3B: Nicotinamide adenine dinucleotide - a hydride transfer coenzyme
Although we are talking here about hydrides acting as nucleophiles and leaving groups,
you already know that literal hydride ions are far too unstable to exist as discreet
intermediates in the organic reactions of living cells (the pKa of H2, the conjugate acid of
hydride, is about 35: a very weak acid, meaning hydride is a very strong base and not a
reasonable species to propose for a biochemical reaction). As was alluded to earlier,
biochemical hydrogenation/dehydrogenation steps require the participation of a
specialized hydride transfer coenzyme. The most important of these is a molecule called
nicotinamide adenine dinucleotide. The full structure of the oxidized form of this
coenzyme, abbreviated NAD+, is shown below, with the active nicotinamide group
colored blue. Because the redox chemistry occurs specifically at the nicotinamide ring (in
blue in the figure below), typically the rest of the molecule is simply designated as an 'R'
group.
O
NH2
NH2
nicotinamide ring
O
N
O
O P O P O
O
HO
N
O
OH
O
N
O
HO
N
N
OH
NAD
NADP+ if phosphorylated here
fig 16
If the hydroxyl group indicated by the arrow is phosphorylated, the coenzyme is called
NADP+. The phosphate is located far from the nicotinamide ring and does not participate
272
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
directly in the hydride transfer function of the cofactor. It is, however, important in a
larger metabolic context: as a general rule, redox enzymes involved in catabolism (the
breakdown of large molecules) typically use the non-phosphorylated coenzyme, while
those involved in anabolism (biosynthesis of large molecules from small precursors) use
the phosphorylated coenzyme.
NAD+ and NADP+ both function in biochemical redox reactions as hydride acceptors:
that is, as oxidizing agents. The reduced forms of the coenzyme, abbreviated NADH and
NADPH, serve as hydride donors: that is, as reducing agents.
oxidized form
hydride acceptor
oxidizing agent
reduced form
hydride donor
reducing agent
H
H H
O
O
NH2
NH2
N
N
R
R
NAD(P)+
NAD(P)H
fig 5
fig 17
To understand how the nicotinamide coenzymes function in hydride transfer, let's look at
a general picture of a reversible, redox conversion from a ketone to a secondary alcohol.
Mechanistically, the reaction we are about to see can be described as a nucleophilic
addition to a carbonyl - a mechanism type we studied in chapter 10 - with the twist that
the nucleophilic species is a hydride ion. At the beginning of the reaction cycle, both the
ketone substrate and the NADH cofactor are bound in the enzyme's active site, and
carbon #4 of the nicotinamide ring is positioned very close to the carbonyl carbon of the
ketone.
NAD(P)H-dependent hydrogenation (reduction) of a ketone
NAD(P)H, H+
OH
O
R
C
NAD(P)+
R
R C H
R
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
273
Chapter 15: Oxidation and reduction
Mechanism:
H A
OH
O
R
NAD(P)H serves as donor of
nucleophilic hydride
C
2
R
H O
H
3
R C H
R
4
N1
5
H
O
NH2
NH2
6
N
R
R
NAD(P)+
NAD(P)H
fig 18, fig 18a
As an enzymatic group transfers a proton to the ketone oxygen, the carbonyl carbon loses
electron density and becomes more electrophilic, and is attacked by a hydride from
NADH. Because carbon #4 of NADH is bound in such close proximity to the
electrophile, this step can occur without generating a free hydride ion intermediate – the
two hydride electrons can be pictured as shifting from one carbon to another. Note the
products of this reaction: the ketone (which accepted a hydride and a proton) has been
reduced to an alcohol, and the NADH cofactor (which donated a hydride) has been
oxidized to NAD+.
The dehydration of an alcohol by NAD+ is simply the reverse of a ketone hydrogenation:
+
NAD(P) -dependent dehydration (oxidation) of an alcohol
OH
R C H
R
274
NAD(P)+
NAD(P)H, H+
O
R
C
R
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Mechanism:
B
O
H
O
R C H
R
NAD(P)+ serves as acceptor of
hydride leaving group
H
R
O
C
H H
R
O
NH2
NH2
N
N
R
R
NAD(P)+
NAD(P)H
fig 19 fig 19a
An enzymatic base positioned above the carbonyl removes a proton, and the electrons in
the O-H bond shift down and push out the hydride, which shifts over to carbon #4 of
NAD+. Note that the same process with a primary alcohol would yield an aldehyde
instead of a ketone.
Exercise 15.4: Draw general mechanisms for:
a) hydrogenation of an imine
b) dehydrogenation of an amine
Exercise 15.5: We just saw that when the nucleophile in a nucleophilic carbonyl addition
step is a hydride ion from NADH, the result is a ketone/aldehyde hydrogenation reaction.
As a review: what kind of reaction step results when the nucleophile in this process is not
a hydride ion but an enolate carbon?
The nicotinamide coenzymes also serve as hydride donors/acceptors in the redox
reactions interconverting carboxylic acid derivatives and aldehydes. Notice that these
reactions can be thought of as nucleophilic acyl substitution reactions (chapter 11) in
which the nucleophile or leaving group is a hydride ion.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
275
Chapter 15: Oxidation and reduction
NAD(P)H-dependent hydrogenation (reduction) of a thioester to an aldehyde:
NAD(P)H, H+
RSH
NAD(P)+
O
R
O
C
R
SR
C
H
Mechanism:
R
O
O
C
R C
SR
O
H A
SR
R
H
+
C
H
hydride nucleophile
H O
H
H
O
NH2
NH2
N
N
R
R
NAD(P)H
NAD(P)+
fig 20 fig 20a
NAD(P)+-dependent dehydrogenation (oxidation) of an aldehyde to a thioester:
RSH
NAD(P)+
O
R
276
C
H
NAD(P)H, H+
O
R
C
SR
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
RSH
Chapter 15: Oxidation and reduction
Mechanism:
R
C
O
O
O
R C
H
SR
R
H
B
C
SR
hydride leaving group
H
RS
H
O
H H
O
NH2
NH2
N
N
R
R
NAD(P)+
NAD(P)H
fig 21 fig 21a
To simplify figures, hydrogenation and dehydrogenation reactions are often drawn with
the role of the coenzyme abbreviated:
H A
OH
O
R
C
R C H
R
+
R
NAD H
NAD+
:B
O
H
O
C
R C H
R
R
NAD+
NAD H
R
+
fig 22
However, it is very important to make sure that you can remember and draw out the full
mechanism, including the role of the coenzyme, in these types of reactions.
Caution! A very common error made by students learning how to draw biochemical
redox mechanisms is to incorrectly show nicotinamide coenzymes acting as acids or
bases. Remember: NADH and NADPH are hydride donors, NOT proton donors. NAD+
and NADP+ are hydride acceptors, NOT proton acceptors.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
277
Chapter 15: Oxidation and reduction
15.3C: Stereochemistry of ketone hydrogenation
It should not surprise you that the stereochemical outcomes of enzymatic hydrogenation /
dehydrogenation steps are very specific. Consider the hydrogenation of an asymmetric
ketone: In the hydrogenation direction, attack by the hydride can occur from either the re
or the si face of an asymmetric ketone (see section 10.1C), leading specifically to the S or
R alcohol.
H A
OH
O
si
NAD H
re
C
H
R1 R2
C R
1
R2
R configuration
coenzyme bound next to si face
(assume R2 is higher priority than R1)
H A
OH
O
si
re
C
H NAD
R1
R1 R2
coenzyme bound next to re face
C
R2
H
S configuration
fig 23
The stereochemical configuration of the product depends on which side of the ketone
substrate the NAD(P)H coenzyme is bound in the active site. Any given enzyme will
catalyze its reaction with one of these two stereochemical outcomes, not both.
Stereochemical considerations apply in the dehydrogenase direction as well: in general,
enzymes specifically catalyze the oxidation of either an R or S alcohol, but not both.
OH
(R)
NAD+
H
C R
1
R2
O
C
R and S alcohols are oxidized by
different enzymes
R1 R2
OH
(S)
R1
C
R2
H
NAD+
fig 24
278
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Exercise 15.6: During an intense workout, lactic acid forms in muscle tissue as the result
of enzymatic reduction of a ketone group in the precursor molecule. (EC 1.1.1.27) It is
the lactate that you can blame for the sore muscles you feel the day after a workout.
OH
OH
O
lactic acid
fig 25
a) Draw a mechanism for this reaction, including the structure of the starting ketone and
the nicotinamide ring section of the cofactor.
b) Which face of the ketone is the coenzyme positioned next to in the active site of the
enzyme?
15.3D: Examples of biochemical carbonyl/imine hydrogenation
Now that we have covered the basics, let's look at some real examples of hydrogenation
and dehydrogenation reactions.
Glycerol phosphate dehydrogenase (EC 1.1.1.8) catalyzes one of the final chemical steps
in the breakdown of fat molecules. The enzyme specifically oxidizes (R)-glycerol
phosphate to dihydroxyacetone phosphate. (S)-glycerol phosphate is not a substrate for
this enzyme. J. Mol. Biol. 2006, 357, 858 (human crystal structure)
enz
NH2
H
O H
H
OP
OH
(R)-glycerol
phosphate
O
O
H H
NH2
NH2
OP
N
R
O
OH
dihydroxyacetone
phosphate
N
R
fig 26
The reverse reaction (catalyzed by the same enzyme) converts dihydroxyacetone
phosphate to (R)-glycerol phosphate, which serves as a starting point for the biosynthesis
of membrane lipid molecules (see section 1.3A).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
279
Chapter 15: Oxidation and reduction
H A
si
H NADP
O re
OH
enzyme in eukaryotes
OH OP
dihydroxyacetone
phosphate
HO H
=
H
OP
OH
OP
OH
(R)-glycerol
phosphate
fig 26a
Exercise 15.7: X-ray crystallography experiments reveal that in the active site of glycerol
phosphate dehydrogenase, a zinc cation (Zn+2) is coordinated to the oxygen atom of the
carbonyl/alcohol group. How does this contribute to catalysis of the reaction?
While the cell membranes of animals, plants, and bacteria are made from lipids with the
R stereochemistry exclusively, archaeal microbes (the so-called 'third kingdom of life)
are distinguished in part by the S stereochemistry of their membranes.
HO H
membrane lipids in animals,
plants, and bacteria
OP
O
OP
OH
(R)-glycerol
phosphate
OH
dihydroxyacetone
phosphate
H OH
membrane lipids in archaea
OP
OH
(S)-glycerol
phosphate
fig 27
Archaea have an enzyme that catalyzes hydrogenation of dihydroxyacetone with the
opposite stereochemistry compared to the analogous enzyme in bacteria and eukaryotes.
This archaeal enzyme was identified and isolated in 1997.
280
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
H A
si
O re
OH
H NADP
enzyme in archaea
OH OP
H OH
=
H
PO
dihydroxyacetone
phosphate
OP
OH
OH
(S)-glycerol
phosphate
fig 28
In a reaction that is relevant to people who enjoy the occasional 'adult beverage', an
NADH-dependent enzyme (EC 1.1.1.1) in brewer's yeast produces ethanol by reducing
acetaldehyde. This is the final step in the process by which yeast ferment glucose to
ethanol.
NADH
O
C
H3C
NAD+
OH
H3C C
H
H
H
acetaldehyde
ethanol
fig 29
The reaction below, which is the final step in the biosynthesis of proline (EC 1.5.1.2), is
an example of an enzymatic reduction of an imine to an amine. J. Mol. Biol. 2005, 354,
91.
R
R
N
N
H2N
H2N
O
H H
CO2
enz
N
H
H H
H
O H
O
H
CO2
H N
H
proline
fig 30
This step in the breakdown of the amino acids glutamate (EC 1.4.1.2) provides an
example of the oxidation of an amine to an imine: Structure 1999, 7, 769.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
281
Chapter 15: Oxidation and reduction
CO2
NAD+
NAD H
H
H2N
CO2
H2N
H
CO2
CO2
H2N
enz
The 'double reduction' reaction below (EC 1.1.1.34) is part of the isoprenoid biosynthetic
pathway, which eventually leads to cholesterol in humans.
2 NADPH
HSCoA
2 NADP+
HO CH3 O
O2C
HO CH3 OH
O2C
SCoA
H
H
fig 32a
In this reaction, a thioester is first reduced to an aldehyde in steps 1a and 1b:
HO CH3 O
HO CH3 O
O2C
SCoA
step 1a
O2C
H
H O
H
step 1b
H A
N
HO CH3 OH
O2C
HSCoA
HO CH3 O
R
+
SCoA
NH2
first NADPH
NADP+
H A
O2C
H
NADP+
H O
H
H
H
+
step 2
NH2
N
second NADPH
R
Then in step 2, the aldehyde is in turn reduced by the same enzyme (and a second
NADPH that enters the active site) to a primary alcohol. This enzyme is inhibited by
282
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
atorvastatin and other members of the statin family of cholesterol-lowering drugs.
Atorvastatin, marketed under the trade name Lipitor by Pfizer, is one of the all-time bestselling prescription medications.
Recall from chapter 11 that carboxylates are not reactive in acyl substitution steps, so it
follows that they cannot be directly reduced to aldehydes by an enzyme in the same way
that thioesters can. However, a carboxylate can be converted to its 'activated' acyl
phosphate form (section x.x), which can then be hydrogenated. An example of this is
found in a two-reaction sequence found in amino acid metabolism (EC 2.7.2.11;
EC1.2.1.41).
ATP
H NH3
O
O
CO2
O
H NH3 NADH
ADP
NAD+
H NH3
O
CO2
CO2
H
OP
first, the carboxylate must be
activated by a kinase enzyme . . .
then the hydride from NADH drives off the
phosphate leaving group in an acyl-substitution step
fig 33
Glyceraldehyde-3-phosphate dehydrogenase (EC 1.2.1.12) , a key enzyme in the
glycolysis pathway, provides an example of the oxidation of an aldehyde to a thioester, in
this case a thioester linkage between the substrate and a cysteine residue in the enzyme's
active site. In the second phase of the reaction, the thioester intermediate is hydrolyzed
to free the carboxylate product.
PO
H
PO
OH
+ NADH
O
O
S
enz
OH
S
O
NAD+
H
enz
S
PO
OH
enz
H
:B
aldehyde is oxidized to thioester
H2O
thioester is hydrolyzed
O
O
PO
OH
HS
enz
fig 34
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
283
Chapter 15: Oxidation and reduction
Exercise 15.8: Below is the final step in the biosynthesis of the amino acid histidine (EC
1.1.1.23). Fill in the species that are indicated with question marks. Proc. Natl. Acad. Sci.
U.S.A. 2002, 99, 1859
N
NH
???
OH
H3N
N
???
NH
O
H3N
O
histidine
histidinol
fig 35
Exercise 15.9: Draw a likely mechanism for the conversion of glucose to sorbitol, a
process that occurs in the liver. Do not abbreviate the nicotinamide ring structure.
O
OH
OH
OH
OH
OH
NAD+
NADH
OH
OH
OH
glucose
OH
OH
OH
sorbitol
fig 35a
15.3E: Reduction of ketones and aldehydes in the laboratory
Although our focus in this book is biological organic reactions, it is interesting to note
that synthetic organic chemists frequently perform hydrogenation reactions in the lab that
are similar in many respects to the NAD(P)H-dependent reactions that we have just
finished studying. A reagent called sodium borohydride (NaBH4) is very commonly
used, often in methanol solvent, to reduce ketones and aldehydes to alcohols. The
reagent is essentially a laboratory equivalent of NADH (or NADPH): it serves as a source
of nucleophilic hydride ions. Sodium borohydride is a selective reagent in the sense that
it will reduce ketones and aldehydes but not carboxylic acid derivatives such as esters
(recall from section 11.2 and 11.3 that the carbonyl carbons of carboxylic acid derivatives
are less potent electrophiles than the carbonyl carbons of ketones and aldehydes). Unlike
the enzymatic hydrogenation reactions we saw earlier, the reduction of asymmetric
ketones with sodium borohydride usually results in a 50:50 racemic mixture of the R and
S enantiomers of the alcohol product.
284
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
a hydride gets transferred to ketone
racemic mixture of alcohols
H
O
Na
CH3
H B
H
H
CH3
H
HO H
OH
+
CH3
methanol solvent
fig 36
Synthetic organic chemists have at their disposal a wide range of other reducing and
oxidizing reagents with varying specificities and properties, many of which you will learn
about if you take a course in laboratory synthesis.
Exercise 15.10: Camphor can be easily reduced by sodium borohydride. However, the
mixture of stereoisomeric alcohols that results is not 50:50.
O
camphor
fig 37
a) Draw the two stereoisomers of the alcohol products of this reaction.
b) Indicate which you would expect to be the major product, and explain your reasoning.
c) Which analytical technique - 1H-NMR, IR, UV, or MS - could best be used to
determine the ratio of the stereoisomers in the product mix? Describe how this analysis
could be accomplished.
Section 15.4: Hydrogenation of alkenes and dehydrogenation of alkanes
We turn next to reactions in which a hydrogen molecule is added to the double bond of
an alkene, forming an alkane - and the reverse, in which H2 is eliminated from an alkane
to form an alkene. Many biochemical reactions of this type involve α,β-unsaturated
thioesters.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
285
Chapter 15: Oxidation and reduction
hydride added
H
R β
S
α
O
H H
O
H
R
hydrogenation
R
dehydrogenation
S
R
H H
proton added
fig 38
15.4A: Alkene hydrogenation
In the cell, alkene hydrogenation most often occurs at the α and β position relative to a
carbonyl. This type of alkene hydrogenation is essentially a conjugate addition (section
11.4) of hydrogen, with a hydride ion (often from NAD(P)H) acting as the nucleophile in
the first step.
NAD(P)H-dependent hydrogenation (reduction) of an α,β-conjugated alkene:
H
R α
O
NAD(P)H, H+
S
β
H H
NAD(P)+
R
O
R
S
R
H H
H
Mechanism:
R
N
NH2
H
H
O
enolate intermediate
H
R α
O
β
H
H H
S
O
H H
R
O
R
R
R
S
H
H A
S
R
H H
fig 39 fig 39a
As part of the fatty acid synthesis pathway, a double bond between the α and β carbons
of a fatty acid is reduced to a single bond by hydrogenation (EC 1.3.1.10). The fatty acid
is attached to an acyl-carrier protein via a thioester linkage (section 11.5A).
286
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
H
NADP+
NADPH, H+
O
O
ACP
ACP
R
H H
R
S
S
H H
H
fig 40
15.4B: Flavin-dependent alkane dehydrogenation
Next, let's consider an alkane dehydrogenation reaction (EC 1.3.99.3) in the fatty acid
degradation pathway. Here, a double bond is introduced between the α and β carbons,
with concurrent loss of a hydride ion and a proton.
different redox coenzyme!
CoA thioester in degradation direction
H H
FAD
O
R
SCoA
H H
FADH2
H
O
R
SCoA
H
fig 41
This reaction is clearly not the reverse of the hydrogenation reaction we just saw from
fatty acid biosynthesis. First of all, you should notice that the thioester linkage is to
coenzyme A rather than acyl carrier protein (ACP). More importantly to this
discussion, while the hydride donor in the biosynthetic hydrogenation reaction is
NADPH, the relevant coenzyme in the catabolic direction is not NAD+ or NADP+ rather, it is a flavin coenzyme.
Flavin adenine dinucleotide (FAD) is composed of three components: the three-ring
flavin system, ribose phosphate, and AMP. An alternate form, which is missing the AMP
component, is called flavin mononucleotide (FMN).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
287
Chapter 15: Oxidation and reduction
flavin
O
O
N
N
N
N
H
N
O
N
CH2
AMP
OH
O
HO
O
H2C O P O P O
ribose phosphate
O
N
O
N
N
H
O
CH2
NH2
HO
N
HO
OH
N
O
HO
H2C O P O
N
O
O
OH
HO
flavin mononucleotide (FMN)
flavin adenine dinucleotide (FAD)
fig 42
The reactive part of the coenzyme is the flavin group, so usually the rest of the molecule
is abbreviated with 'R'.
FAD and FMN are the oxidized form of flavin. The reduced (hydrogenated) forms of
these cofactors are abbreviated FADH2 and FMNH2.
O
N
N
H
N
N
R
FAD or FMN
oxidized flavin
reducing agent
O
H
N
O
N
N
R
H
N
H
O
FADH2 or FMNH2
reduced flavin
oxidizing agent
fig 43
The flavin coenzymes are synthesized in humans from riboflavin (vitamin B2), which we
obtain from our diet (the structure of riboflavin is the same as that of FMN, except that
riboflavin lacks the phosphate group). Notice the extended conjugated π system in the
three fused rings: the flavin system absorbs light in the visible wavelengths and has a
distinctive deep yellow color - it is riboflavin, and to some extent FAD and FMN, that
give urine its color.
Like the nicotinamide coenzymes, flavin serves as a hydride donor or acceptor. FAD and
FMN are able to accept a hydride ion (and a proton), and FADH2 and FMNH2 in turn can
serve as hydride donors in hydrogenation reactions.
288
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
hydride delivered from biomolecule being oxidized
hydride delivered to biomolecule being reduced
H
H
O
N
N
R
N
N
O
H
N
O
N
N
R
H
N
H A
H
O
B
reduced flavin
hydride donor
reducing agent
oxidized flavin
hydride acceptor
oxidizing agent
fig 44
Below is a general mechanism for the dehydration of an alkene at the α,β position notice that it is mechanistically an E1cb elimination of H2.
Flavin-dependent α,β dehydrogenation (oxidation) of an alkane:
H H
FAD
O
R
SCoA
H H
FADH2
H
O
R
SCoA
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
289
Chapter 15: Oxidation and reduction
Mechanism:
H H
H H
O
R
O
R
R
R
H
H H
O
N
B
N
N
N
R
H
O
R
H
O
H A
flavin
(oxidized form)
R
H
H
O
N
N
N
N
R
H
H
O
flavin
(reduced form)
fig 45 fig 45a
In many enzymatic reactions in which FADH2 acts as the reducing agent, the reaction
cycle is completed when FAD, rather than being released from the active site, is recycled
back to FADH2 with the concomitant oxidation of NADH.
290
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
R
R
N
N
NH2
NH2
H
H
H O
O
NAD+
NADH
H
O
N
N
N
N
H
N
O
N
N
R
H
R
H A
FAD
O
N
H
O
FADH2
fig 51g
Hydride ion transfer with flavin or nicotinamide coenzymes is a two electron redox
process. However, unlike the nicotinamide cofactors, flavins are also able to function in
single electron transfer (radical) mechanisms. We will come back to this idea briefly in
the chapter 16.
Exercise 15.11: Fumarate is formed in an alkane dehydrogenation reaction which is part
of the citric acid cycle:
FAD
?
FADH2
H
O
O
O
O
H
fumarate
fig 46
a) Predict the structure of the starting substrate in this reaction
b) Draw the structure of the enolate intermediate
Exercise 15.12: Flavin can serve as the hydride donor in some hydrogenation reactions.
Degradation of the RNA base uracil begins with saturation of the conjugated alkene
group by a flavin-dependent hydrogenase enzyme (EC 1.3.1.2). Predict the product of
this step, and draw a likely mechanism for the reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
291
Chapter 15: Oxidation and reduction
O
H
FAD
H
N
O
FADH2
N
H
?
alkene hydrogenation
H
fig 47
Section 15.5: Monitoring hydrogenation and dehydrogenation reactions by UV
spectroscopy
In order to study any enzyme-catalyzed reaction, a researcher must have available some
sort of test, or assay, in order to observe and measure the reaction's progress and measure
its rate. In many cases, an assay simply involves running the reaction for a specified
length of time, then isolating and quantifying the product using a separation technique
such as high performance liquid chromatography (HPLC) or gas chromatography (GC).
This type of assay can be extremely time-consuming, however, so it is to the researcher's
great advantage if a more convenient assay can be found.
Redox reactions in which a nicotinamide coenzyme participates as a hydride donor or
acceptor are generally quite convenient to assay. In fact, the progress of these reactions
can usually be observed in real time, meaning that the researcher doesn't need to stop the
reaction in order to see how far it has progressed. NADPH and NADH have distinctive
n-π* UV absorbance bands centered at 340 nm, with a molar absorptivity of 6290 M-1
cm-1 (section 4.4). The oxidized coenzymes NADP+ and NAD+ do not absorb at this
wavelength.
λmax = 340 nm
NAD(P)H
NAD(P)+
292
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
fig 48
Therefore, the course of a hydrogenation reaction, in which NAD(P)H is converted to
NAD(P)+, can be observed in real time if it is run in a quartz cuvette in a UV
spectrometer. By observing the decrease in absorbance at 340 nm, the researcher can
calculate how much NAD(P)H has been oxidized to NAD(P)+ at any given time point,
and this number is the molar equivalent of the amount of organic substrate that has been
reduced:
no absorbance at 340 nm
absorbs at 340 nm
oxidized substrate +
reduced product
NAD(P)H
+
NAD(P)+
fig 48a
Likewise, a NAD+-dependent dehydrogenase reaction can be followed in real time by
monitoring the increase in absorbance at 340 nm as NAD+ is converted to NADH.
Exercise 15.13: You are observing the progress of the (R)-glycerol phosphate
dehydrogenase reaction shown in the figure above. You run the reaction in a quartz
cuvette (path length 1 cm) in a total solution volume of 1 mL. You start with 200 µM
substrate and 100 µM NAD+ in solution, zero the UV spectrophotometer, then add the
enzyme to start the reaction. After 5 minutes, the A340 reading has climbed from 0.000 to
0.096. At this time point:
a) How many moles of substrate have been oxidized?
b) What is the solution concentration of NADP+?
c) The enzyme has a mass of 25 kilodaltons (25,000 g/mol).You added 5 µL of a 2 ng/µL
solution of pure enzyme to start the reaction. How many reactions does each enzyme
molecule catalyze, on average, per second? (This number is referred to by biochemists as
the 'turnover number').
Section 15.6: Redox reactions of thiols and disulfides
A disulfide bond is a sulfur-sulfur bond, usually formed from two free thiol groups.
reduced form
R
R
SH
SH
two thiols
(dithiol)
oxidized form
R S
R
S
disulfide
fig 49
The interconversion between dithiol and disulfide groups is a redox reaction: the free
dithiol form is in the reduced state, and the disulfide form is in the oxidized state. Notice
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
293
Chapter 15: Oxidation and reduction
that in the oxidized (disulfide) state, each sulfur atom has lost a bond to hydrogen and
gained a bond to sulfur.
As you should recall from your Biology courses, disulfide bonds between cysteine
residues are an integral component of the three-dimensional structure of many
extracellular proteins and signaling peptides.
HS
HS
reduced protein (free cysteines)
S
S
oxidized protein (disulfide)
fig 49
fig 49a
A thiol-containing coenzyme called glutathione is integrally involved in many thioldisulfide redox processes (recall that glutathione was a main player in this chapter's
introductory story about concussion research). In its reduced (thiol) form, glutathione is
abbreviated 'GSH'. In its oxidized form, glutathione exists as a dimer of two molecules
linked by a disulfide group, and is abbreviated 'GSSG'.
294
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
NH3
H
N
O2C
NH3
O
N
CO2
H
N
O2C
N
CO2
H
O
H
O
O
S
HS
glutathione (GSH)
reduced state
S
H
O
O2C
N
N
NH3
H
CO2
O
glutathione dimer (GSSG)
oxidized state
fig 50
Disulfide bonds and free thiol groups in both proteins and smaller organic molecules like
glutathione can 'trade places' through a disulfide exchange reaction. This process is
essentially a combination of two direct displacement (SN2-like) events, with sulfur atoms
acting as nucleophile, electrophile and leaving group.
Disulfide exchange reaction
HS
R1
+
HS
S
R3
R4
HS
+
S
S
R2
S
R1
R3
HS
R2
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
R4
295
Chapter 15: Oxidation and reduction
Mechanism:
H A
R1
B:
S
S
H
R1
R3
S
H A
S
R3
H
R4
HS
S
:B
HS
S
R2
R4
R2
R1
S
HS
+
S
R3
HS
R4
R2
fig 51
In eukaryotes, the cysteine side chains of intracellular (inside the cell) proteins are almost
always in the free thiol (reduced) state due to the high concentration of reduced
glutathione (GSH) in the intracellular environment. A disulfide bond in an intracellular
protein will be rapidly reduced in a disulfide exchange reaction with excess glutathione.
protein
protein
G
S
S
+
G
HS
HS
HS
HS
+
protein
fig 51
296
S
G
G
protein
S
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
The interconversion of free thiols and disulfides is also mediated by flavin in some
enzymes.
Flavin-mediated reduction of a protein disulfide bond
protein
protein
S
H
N
H A
O
N
N
N
R
H
FADH2
S
protein
protein
HS
S
B:
H
H
O
N
O
N
N
N
H
H A
O
R
:B
protein
protein
SH
HS
O
N
N
N
N
H
O
R
FAD
fig51c
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
297
Chapter 15: Oxidation and reduction
Flavin-mediated oxidation of a protein disulfide bond
protein
protein
HS
S
B:
A
protein
protein
H
S
S
H
O
N
N
R
N
N
H
H
N
O
N
:B
H
O
N
N
H
O
R
H A
FAD
protein
protein
S
H
N
S
O
N
N
N
R
H
H
O
FADH2
fig 51h
As was stated earlier, a high intracellular concentration of reduced glutathione (GSH)
serves to maintain proteins in the free thiol (reduced) state. An enzyme called
glutathione reductase catalyzes the reduction of GSSG in a flavin-mediated process, with
NADH acting as the ultimate hydride donor.
298
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Gluthione reductase reaction:
NADH
G
NAD+
S
G
G
G
HS
SH
S
[FADH2]
glutathione
(oxidized)
glutathione
(reduced)
fig 51a
The mechanism for this and other similar reactions is not yet completely understood, but
evidence points to an initial thiol-disulfide exchange reaction with a pair of cysteines
from the enzyme, (phase 1 below) followed by flavin-dependent reduction of the
cysteine-cysteine disulfide (phase 2). Finally, (phase 3) FAD is reduced back to FADH2
by NADH. Frey and Hegeman, Enzymatic Reaction Mechanisms, p. 699
Phase 1: thiol-disulfide exchange (see earlier figure for mechanism):
enz
enz
GSSG
2 GSH
enz
enz
SH HS
S
enzyme
(reduced)
enzyme
(oxidized)
S
fig 51b
Phase 2: Reduction of protein disulfide by FADH2 (see earlier figure for mechanism)
FADH2
FAD
S
enz
enz
enz
enz
SH
S
HS
fig 51c
Phase 3: regeneration of FADH2 by NADH (see section 15.4B for mechanism)
NADH, H+
FAD
NAD+
FADH2
fig 51d
In the biochemistry lab, proteins are often maintained in their reduced (free thiol) state by
incubation in buffer containing an excess concentration of β-mercaptoethanol (BME) or
dithiothreitol (DTT). These reducing agents function in a manner similar to that of GSH,
except that DTT, because it has two thiol groups, can form an intramolecular disulfide in
its oxidized form.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
299
Chapter 15: Oxidation and reduction
OH
HO
SH
SH
HS
OH
β-mercaptoethanol
(BME)
dithiothreitol
(DTT)
fig 52
Exercise 15.14: Draw structures of the oxidized (disulfide) forms of BME and DTT.
Section 15.7: Flavin-dependent monooxygenase reactions: hydroxylation,
epoxidation, and the Baeyer-Villiger oxidation
Up to now, the redox reaction examples we have seen have all been either
hydrogenation/dehydrogenation transformations or interconversions between free thiols
and disulfides. However, there are many important redox reactions in biological
chemistry which do not fall under either of these descriptions. Oxygenase enzymes
catalyze the insertion of one or two oxygen atoms from molecular oxygen (O2) into an
organic substrate molecule. Enzymes which insert a single oxygen atom are called
monooxygenases. Below are two examples of biochemical transformations catalyzed by
monooxygenase enzymes: one is a hydroxylation, the other is an epoxidation (an epoxide
functional group is composed of a three-membered carbon-carbon-oxygen ring - epoxides
are somewhat rare in biological organic chemistry but are very common and useful
intermediates in laboratory organic synthesis).
HO
H2N
H2N
O
O
H3N
H3N
CO2
R
O
CO2
R
fig 56
Dioxygenase enzymes insert both oxygen atoms from O2 into the substrate, and usually
involve cleavage of an aromatic ring. Below is an example of a dioxygenase reaction,
catalyzed by catechol dioxygenase:
300
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
OH
OH
O
O
HO
OH
In the reduction direction, reductases remove oxygen atoms, or sometimes other
electronegative heteratoms such as nitrogen or halides. For example, DNA
deoxyribonucleosides are converted from their corresponding RNA ribonucleosides by
the action of reductase enzymes:
Base
PPO
Base
PPO
O
O
HO
ribonucleotide
reductase
OH
ribonucelotide diphosphate
(RNA)
HO
deoxyribonucleotide diphosphate
(DNA)
fig 58
Many oxygenase and reductase reactions involve the participation of enzyme-bound
transition metals - such as iron or copper - and the mechanistic details of these reactions
are outside the scope of our discussion. A variety of biochemical monooxygenase
reactions, however, involve flavin as a redox cofactor, and we do have sufficient
background knowledge at this point to understand these mechanisms. In flavin-dependent
monooxygenase reactions, the key intermediate species is flavin hydroperoxide.
The term 'peroxide' refers to a functional group characterized by an oxygen-oxygen single
bond. The simplest peroxide is hydrogen peroxide (HOOH) about which we will have
more to say below. In flavin hydroperoxide, the peroxide group is linked to one of the
carbons of the reactive triple-ring system of the coenzyme. A possible mechanism for the
formation of flavin peroxide from FADH2 and molecular oxygen is shown below.
Silverman, R.B. The Organic Chemistry of Enzyme-Catalyzed Reactions, p. 121-122,
Scheme 3.33. 2000, Academic Press, San Diego.
Mechanism for the formation of flavin hydroperoxide:
singlet oxygen
A
H
O O
H
H
O
N
N
N
N
O
R
H
:B
FADH2
N
H
N
OH
O
O
N
N
H
O
R
flavin hydroperoxide
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
301
Chapter 15: Oxidation and reduction
fig 59a
(Note: Implicit in this mechanism is that the molecular oxygen first undergoes spin
inversion from the triplet state to the higher energy 'singlet' state. You may recall from
your general chemistry course that molecular oxygen exists in two states: 'singlet' oxygen
has a double bond and no unpaired electrons, while 'triplet' oxygen has a single O-O bond
and two unpaired electrons - a kind of 'double radical'. Molecular orbital theory - and
experimental evidence - show that the triplet state is lower in energy.
O O
O O
triplet oxygen
(ground state)
singlet oxygen
(excited state)
fig 60
The mechanism shown above is one proposed mechanism, another proposal involves
triplet oxygen reacting with flavin in a series of radical-intermediate, single-electron
steps.)
Flavin hydroperoxide can be thought of as an activated form of molecular oxygen.
Peroxides in general are potent oxidizing agents, because the oxygen-oxygen single
bond is quite weak: only 138 kJ/mole, compared to 339 kJ/mol for a carbon-carbon bond,
and 351 kJ/mol for a carbon-oxygen bond. When the 'outer' oxygen of flavin
hydroperoxide (red in our figure above) comes into close proximity to the π-bonded
electrons of an alkene or aromatic group, the O-O bond will break, leaving an empty
orbital on the outer oxygen to be filled by the π electrons - thus, a new carbon-oxygen
bond is formed. This is what is happening in step 1 of a reaction in the tryptophan
degradation pathway catalyzed by kynurenine 3-monooxygenase. Step 2 completes what
is, mechanistically speaking, an electrophilic aromatic substitution reaction (section 14.4)
with an peroxide oxygen electrophile.
302
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Mechanism for the flavin hydroperoxide-dependent hydroxylation of kynurenine:
H
O
OH
N
N
N
N
H
O
R
A
H
H
N
N
OH
O
O
H2N
N
N
R
flavin hydroperoxide
H
O
step 1
O
H3N
CO2
B:
kynurenine
HO
H
H2N
flavin hydroperoxide
O
step 2
HO
H3N
CO2
H2N
O
H3N
CO2
3-hydroxykynurenine
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
303
Chapter 15: Oxidation and reduction
fig 61a
B:
H
N
N
H A
O
OH
O
N
N
N
H
step 3
O
N
N
N
R
R
flavin hydroxide
FAD
H
O
NADH
step 4
NAD+
O
H
N
N
N
N
H
R
H
O
FADH2
fig 61
Elimination of water from the hydroxyflavin intermediate then leads to formation of FAD
(step 3), which is subsequently reduced back to FADH2 by NADH (step 4).
The N-hydroxylation reaction below, which is part of the of the biosynthetic pathway of
an iron-binding molecule in the pathogenic bacterium Pseudomonas aeruginosa, is
mechanistically similar to the C-hydroxylation reaction we just saw, except that the
nucleophile is an amine nitrogen.
OH
NH2
H3N
CO2
ornithine
NADPH
O2
NADP+
H2O
[FAD]
NH
H3N
CO2
N-hydroxyornithine
fig 62b
Exercise 15.15: Draw arrows for the N-O bond-forming step in the ornithine
hydroxylation reaction above.
304
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Epoxides, characterized by a three-membered ring composed of two carbons and one
oxygen, are a very common and useful functional group employed in synthetic organic
chemistry. Although rare, there are some interesting epoxide-forming reactions in
biochemical pathways, catalyzed by flavin-dependent monooxygenase enzymes.
In a key step in the biosynthesis of cholesterol and other steroid compounds, an alkene is
converted to an epoxide in a precursor molecule called squalene. Flavin hydroperoxide
also serves as the direct oxidizing agent in this step:
Mechanism for the flavin-hydroperoxide-dependent epoxidation of squalene:
R
A
H
R
O
H
O
H
N
N
R
O
squalene
O
N
N
B:
O
H
oxidosqualene
H
O
R
flavin hydroperoxide
squalene
fig 62
Oxidosqualene goes on to cyclize to lanosterol in a complex and fascinating electrophilic
reaction which we discussed in section 14.5.
Epoxidation reactions have a parallel in the synthetic organic laboratory, and in fact are
very important tools in organic synthesis. In laboratory epoxidations, peroxyacids are
the counterpart to flavin hydroperoxide in biochemical epoxidations. metachloroperoxybenxoic acid (MCPBA) is a commonly used peroxyacid.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
305
Chapter 15: Oxidation and reduction
O
O
R
Cl
O
O
OH
OH
meta-chloroperoxybenzoic acid
(MCPBA)
a peroxyacid
fig 62d
The Baeyer-Villiger oxidation, in which a ketone is converted to an ester through
treatment with a peroxide reagent, is an extremely useful laboratory organic synthesis
reaction discovered in the late 19th century. Recently, many biochemical examples of
Baeyer-Villiger oxidations have been discovered: the reaction below, for example, is
catalyzed by a monooxygenase in a thermophilic bacterium:
(Proc. Natl. Acad. Sci. U.S.A. 2004, 101, 13157)
A Baeyer-Villiger oxidation:
O
NADPH
O2
NADP +
H 2O
O
oxygen atom has inserted
between two carbons
O
[FAD]
Mechanism:
B:
nucleophilic addition of
flavin peroxide to ketone
flavin
H O
flavin
O
A
H
O
flavin
OH
O
O
O
step 2
step 1
O
O
alkyl group shifts from
carbon to oxygen
fig 62a
The Baeyer-Villiger mechanism is differs significantly from the hydroxylation reactions
we saw earlier, although flavin hydroperoxide (abbreviated in the above figure) still plays
a key role. Here, the peroxide oxygen is a nucleophile, rather than an electrophile,
attacking the ketone carbonyl in step 1. Step 2 is a rearrangement, similar in many ways
to the hydride and alkyl shifts we learned about in section 14.5. The electrons in the red
bond in the figure shift over one atom: from the carbonyl carbon to the outer peroxide
oxygen. The end result is that an oxygen atom, from O2 via flavin hydroperoxide, has
been inserted between the carbonyl carbon and a neighboring methylene (CH2) carbon,
forming an ester.
306
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Note that in the reaction mechanism above, the ketone substrate is asymmetric: on one
side of the carbonyl there is a benzyl group (CH2-phenyl), and on the other side a methyl
group. Note also that it is the benzyl group, not the methyl, that shifts in step 2 of the
mechanism. For reasons that are not yet well understood, in Baeyer-Villiger reactions the
alkyl group with higher carbocation stability has a higher migratory aptitude: in other
words, it has a lower energy barrier for the shifting step.
Exercise 15.16: Draw the product of a hypothetical Baeyer-Villiger reaction involving the
same substrate as the above figure, in which the methyl rather than the benzyl group
shifts.
Exercise 15.17: Draw the likely major product of a hypothetical Baeyer Villiger reaction
starting with 2-methylcyclopentanone as the substrate. Take into account the idea of
migratory aptitude.
Below is another example of a Baeyer-Villiger reaction in which a cyclic ketone is
oxidized to a lactone (cyclic ester). Notice that oxygen insertion expands the ring from 6
to 7 atoms. This is the third-to-last step in the biosynthesis of the anti-cancer agent
mithromycin in some bacterial species (ACS Chem. Biol. 2013, 8, 2466).
H
R1O
O
CH3
OH
NADP+
H2O
NADPH
O2
CH3
H3C
OH
O
O
OR2
OH
O
CH3
OH
CH3
O
H3C
[FAD]
OH
H
R1O
O
OH
O
O
O
OR2
(R1 and R2 are triglycerides)
fig 62c
Yet another variety of flavin-dependent monooxygenase, which bears some mechanistic
similarity to the Baeyer-Villiger oxidation, is the decarboxylative reaction below from
biosynthesis of the plant hormone auxin: (J. Biol. Chem. 2013, 288, 1448)
O
CO2
N
H
NADPH
O2
CO2
NADP+
H2O
CO2
N
H
[FAD]
auxin
Exercise 15.18: Propose a mechanism for the above reaction, starting with flavin
hydroperoxide.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
307
Chapter 15: Oxidation and reduction
Section 15.8: Hydrogen peroxide as a dangerous Reactive Oxygen Species
We get our energy from the oxidation of organic molecules such as fat and carbohydrates,
as electrons from these reduced compounds are transferred to molecular oxygen, thereby
reducing it to water. Reducing O2, however, turns out to be a hazardous activity: harmful
side products called reactive oxygen species (ROS) are inevitably formed in the process.
Recall from the story introducing this chapter that ROS appear to play an important role
in the damage that occurs to the brain immediately after a concussion.
Hydrogen peroxide, HOOH, is an ROS. Recall peroxides are potent oxidizing agents due
to the weakness of the O-O single bond. It is this same weak bond that causes hydrogen
peroxide to be dangerous when produced in our bodies, as it can react spontaneously with
oxygen or nitrogen nucleophiles and π bonds.
Peroxide formed as a by-product of our metabolism is particularly harmful when it
oxidizes DNA bases. In just one of many known examples of oxidative damage, the
DNA base cytosine is oxidized to thymine glycol in the presence of hydrogen peroxide.
Although mechanistic details for reactions such as these are not yet well understood, one
possibility is electrophilic addition:
NH2
H
O
O
NH2
H
O
OH
N
N
N
DNA
O
NH2
N
OH
DNA
OH
N
O
N
OH
DNA
thymine glycol
cytosine
fig 53
Our bodies have evolved ways to dispose of the harmful reactive oxygen species that are
continuously being formed (the only way to stop the production of ROS is to stop
breathing oxygen!). Glutathione peroxidase is a remarkable enzyme in that its active site
contains selanocysteine, a modified cysteine residue in which the side chain sulfur is
replaced by selenium (selenium is very toxic, but we do need a very small amount of it
in our diet). Look at a periodic table: selenium is below oxygen and sulfur in the same
column. If you think back to the vertical periodic trends in nucleophilicity (section
8.2B), you'll recall that just as a thiol is a better nucleophile than an alcohol, a selanol
(RSeH) is even more nucleophilic than a thiol. Moreover, the vertical periodic trend in
acidity (section 7.3A) tells us that a selenol should be more acidic than a thiol - in fact,
the pKa of a selenocysteine is about 5.5, meaning that it is mostly in its deprotonated
state at physiological pH, making it even more nucleophilic.
308
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
Glutathione peroxidase very efficiently catalyzes the reduction of hydrogen peroxide to
water and the oxidation of glutathione (GSH) to GSSG, beginning with nucleophilic
attack by the enzymatic selanocysteine on a peroxide oxygen. The intermediates in this
process are shown below: each step can be thought of as a concerted nucleophilic
displacement similar to those that take place in a disulfide exchange reaction.
GSH
H2O
enz
+
Se
H
O
O
H
H2O
enz
enz
Se
OH
Se
SG
GSH
GSSG
enz
Se
fig 54
Eur. J. Biochem. 1983, 133, 51
Biochem. et Biophys. Acta 2009, 1790, 1486
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
309
Chapter 15: Oxidation and reduction
Key learning objectives for this chapter
Before moving on to the next chapter, you should be able to:
Recognize when an organic molecule is being oxidized or reduced, and distinguish
between redox and non-redox organic reactions.
Draw complete mechanisms for the following reaction types, including the structure of
the reactive part of the redox coenzyme (it is strongly recommended that you commit to
memory the structures of the reactive parts of the nicotinamide and flavin coenzymes).
oxidation of an alcohol to an aldehyde or ketone
oxidation of an amine to an imine
oxidation of an aldehyde to a carboxylic acid derivative (usually a thioester or
carboxylate)
oxidation of an alkane to an alkene at the α,β position relative to a carbonyl or
imine
reduction of an aldehyde or ketone to an alcohol
reduction of an imine to an amine
reduction of a carboxylic acid derivative to an aldehyde
reduction of an α,β - conjugated alkene to an alkane
oxidation of two thiol groups to a disulfide in a disulfide-exchange type reaction
reduction of a disulfide group by flavin
flavin hydroperoxide-dependent hydroxylation, epoxidation, and Baeyer-Villiger
reactions
reduction of FAD (or FMN) to FADH2 (or FMNH2) by NAD(P)H.
spontaneous oxidation of an alkene group in a biomolecule by hydrogen peroxide
reduction of hydrogen peroxide by glutathione peroxidase
In addition, you should be able to draw complete mechanisms for hydrogenationdehydrogenation and disulfide exchange reactions that we have not yet seen specific
examples of, based on your understanding of the chemistry involved in these reaction
types and organic reaction patterns in general. Several exercises and end-of-chapter
problems provide opportunities practice with inferring and drawing mechanisms of less
familiar redox reactions.
Given a multistep pathway diagram, you should be able to recognize the transformations
taking place and fill in missing intermediate compounds or reagents (problem 15.5 is an
example of this type).
310
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
You should be working on gaining proficiency at solving multi-step pathway elucidation
problems, such as those at the end of this chapter's problem section.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
311
Chapter 15: Oxidation and reduction
Problems
P15.1: Show a mechanism for each of the redox reactions below. Do not abbreviate the
reactive parts of the redox coenzyme.
a)
HO
NADH, H+
O
NAD+
HO
O
adenine
adenine
O
HO
OH
OH
adenosine
b)
H2O
NAD+
NH3
O
NADH, H+
O
H
NH3
O
O
O
O
O
c)
NADH, H+
NH3
O
NAD+
Pi
NH3
O
CO2
CO2
H
OP
d)
O
NADPH, H+
O
NADP+
OH
O
ACP
ACP
S
S
e)
FAD
O
O2C
312
SCoA
FADH2
O
O2C
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
SCoA
Chapter 15: Oxidation and reduction
f)
O
HSCoA
NAD+
O
H
NADH, H+
O
O
CoAS
O
O
g) (regeneration of reduced flavin by NADH):
NADH, H+
NAD+
FADH2
FAD
h) In reaction (a), near which face of the substrate is the cofactor bound in the active site?
i) ) In reaction (d), near which face of the product is the cofactor bound in the active site?
P15.2: In the transformation below, the side chain of aspartate is altered, but the main
peptide chain is not affected. Show the most probable structure of species A and B.
O
ATP
O
O
H3N
H+
NADPH
ADP
Pi
NADP+
A
B
O
P15.3: Propose complete mechanisms for the following reactions.
a) (from lysine biosynthesis)
OH
O2C
N
2 H+
NADPH
CO2
H2O
NADP+
O2C
N
CO2
(
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
313
Chapter 15: Oxidation and reduction
b) (from guanosine ribonucleotide biosynthesis - the mechanism involves a covalent
cysteine-linked enzyme-substrate intermediate)
H2O
NAD+
O
N
N
NH
N
N
N
R
O
NADH, H+
R
HS
NH
N
H
O
enz
P15.4: The first step in the lysine degradation pathway is a reductive condensation with
α-ketoglutarate to form an intermediate called saccharopine.
a) Propose a mechanism for this transformation.
lysine
NADPH, H+
O
O2C
H2O
NADP+
CO2
NH3
CO2
O2C
N
H
CO2
saccharopine
α-ketoglutarate
b) Saccharopine (see part (a) above) is then broken up to yield glutamate and a second
product that contains an aldehyde group. Predict the structure of this second product, and
propose a likely mechanism for the reaction which involves and imine intermediate.
NAD+
NH3
CO2
O2C
N
H
CO2
H2O
NADH, H+
imine
intemediate
saccharopine
(
314
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
glutamate
?
Chapter 15: Oxidation and reduction
P15.5: Predict the structures of species A and B in the pathway below.
2 H2O
O
HSCoA
NAD+
NADH, H+
NADH, H+
B
A
SCoA
CoASH
NAD+
C
CO2
O
CoAS
P15.6: Bilirubin, the molecule responsible for the yellowish color of bruises, is formed
from the NADPH-dependent hydrogenation of a double bond in biliverdin (EC 1.3.1.24),
which is a product of heme degradation (heme is an iron-containing coenzyme in the
oxygen-carrying blood protein hemoglobin). Draw a likely mechanism for this reaction.
NADPH, H+
N
H
NAD+
N
N
H
N
H
bilirubin
biliverdin
P15.7: Draw a likely mechanism for each of the reactions below.
a) From the oxidation of polyunsaturated fatty acids: (J. Biol. Chem 2005, 280, 3068)
NADPH, H+
NADP+
CO2
CO2
b) From nucleic acid biosynthesis:
OH
NADPH, H+
O2C
N
CO2
H2O
NADP+
O2C
N
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
CO2
315
Chapter 15: Oxidation and reduction
d) ChemBioChem 2013, 14, 1998, Scheme 2
O
+
R
FADH 2
HS-ACP
FAD
2H2O
S-ACP
O
O
H 2N
NH 2 O
OH
R
+
OH
OH
O
P15.8: Predict the structures of pathway intermediates A, B, and C:
OP
NADP+
HO
NADPH, H+
NADP+
H2O
NADPH, H+
O
HO
A
OH
B
C
OH
glucose
CO2
OP
HO
OH
O
OH
ribulose-5-phosphate
P15.9: An enzyme called DsbA is responsible for the formation of disulfide bonds in
bacterial proteins. The process - which can be thought of as a 'disulfide exchange',
involves the cleavage of a disulfide bond between two active site cysteines in DsbA. It is
accomplished through two successive SN2 displacements.
H
H
H
S
S
S
S
2 SN2 steps
S
DsbA
S
S
substrate protein
Biochem 39, 6732
316
S
+
+
substrate protein
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
DsbA
Chapter 15: Oxidation and reduction
DsbA is then returned to it's starting (disulfide) state through a second disulfide exchange
reaction with another protein called DsbB:
H
H
S
S
S
S
intermediate being
studied
step 3
DsbA
DsbB
step 4
H
S
S
H
S
DsbA
S
DsbB
Scientists were interested in studying the intermediate species formed in step 3, but found
that it is very short-lived and difficult to isolate. In order to address this problem, they ran
the reaction with a synthetic analog of DsbB that contained an unnatural bromoalanine
amino acid in place of one of the active site cysteines.
a) Draw a complete mechanism for the disulfide exchange reaction between DsbA and
DsbB.
b) Show how the bromoalanine-containing DsbB analog allowed for the isolation of an
intermediate that resembles the true, short-lived intermediate.
(J. Am. Chem. Soc. 2004, 126, 15060;
2001, 64, 597).
P15.10:
J. Nat. Prod.
a) In chapter 16 we will learn how ascorbate (vitamin C) acts as a 'radical scavenger'
antioxidant to protect our cells from damage by free radical species. When ascorbate
scavenges a radical, it ends up being converted to dehydroascorbate . One possible
metabolic fate of dehydroascorbate is to be recycled back to ascorbate through an
enzyme-free reaction with glutathione. Biochemistry 1999, 38, 268.
HO H
O
O
2 GSH
H2O
GSSG
HO H
HO
O
OH
OH OH
dehydroascorbate
(hemiacetal form)
O
O
OH
O
ascorbate
probfig10a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
317
Chapter 15: Oxidation and reduction
Suggest a likely mechanism for the enzyme-free reaction involving an SN1-like
dehydration step.
b) In the introduction to chapter 16, we will learn that most animals - but not humans are able to synthesize their own ascorbate. Humans cannot synthesize vitamin C because
we lack the enzyme for the final step in the biosynthetic pathway, gulonolactone oxidase:
HO
H
O
O
HO
O2
HO
H2O2
H
O
HO
O
HO
[FAD]
O
OH
O
ascorbate
gulonolactone
This enzyme uses FAD as an oxidizing agent, and FADH2 is oxidized back to FAD at the
end of the catalytic cycle by molecular oxygen, with hydrogen peroxide as a side product.
Draw out a likely mechanism showing how gulonolactone is converted to ascorbate, and
how FAD is regenerated.
c) Artemisinin is a naturally occurring compound with demonstrated antimalarial
properties. It is thought to act by depleting the malaria-causing microbe's store of reduced
flavin, thus disrupting the redox balance. The relevant reaction is shown below:
H
FADH2
O
O
O
H
FAD
HO
O
H
O
O
H
CO2
O
artemisinin
Draw mechanistic arrows for the step as shown above. (Molecules 2010, 15, 1705)
d) Draw the mechanistic step in the reaction below in which the C-O bond indicated by
the arrow in the figure below is formed.
O
NADPH
O2
NADP+
H2O
O
O
[FAD]
P15.11: Methanogens are a class of microorganisms in the domain archaea which inhabit
a diversity of anaerobic (oxygen-lacking) environments, from the intestines of humans, to
318
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
swamp mud, to the base of deep sea hot water vents. They obtain energy by reducing
carbon dioxide to methane:
CO2 + 4H2 → CH4 + 2 H2O + energy
Methanogenesis', like the oxidation of glucose in animals, is not accomplished in a single
reaction - it requires a long series of enzymatic steps, and involves the participation of
several unique coenzymes (if you are interested in learning more, see FEMS Microbiol.
Rev. 1999, 23, 13 for a detailed review of the enzymatic reactions of methanogenesis).
The oxidation of methane to carbon dioxide when you burn natural gas for heating your
house is obviously an exergonic process. How then is it possible that reducing carbon
dioxide to methane could also be exergonic? Explain.
Multistep pathway prediction problems
In chapters 12 and 13, we were introduced to the challenge of mapping out potential
multi-step transformations using the retrosynthesis approach, where we start with the
more complex molecule and take it apart step be step. These problems have necessarily
involved hypothetical, generalized transformations, because many of the reaction types
involved in actual biochemical transformations were as-yet unfamiliar to us.
We have now reached a point in our study of organic reactivity where we can look at an
actual metabolic pathway and probably recognize most of the reactions taking place - so
were are ready to try our hand at mapping out real metabolic pathways.
Before we dive in, you may want to go back to the retrosynthesis interchapter and review
the key elements of the retrosynthetic approach. You may (or may not) also need a little
more practice with some simpler, shorter, hypothetical problems, similar to what we
worked on in the last three chapters but incorporating some of the redox reactions that we
have just finished learning about. These are provided in problems 15.12 and 15.13
below.
As before, your job is to draw out a 'pathway diagram' for each transformation, using the
'arrow in - arrow out' convention to indicate the role of other necessary participants in
each reaction, such as ATP, NADH, water, or another organic molecule. An example for
a simple two-step pathway is provided in problem 15.12 below. A three-step pathway
would of course show two intermediate compounds.
Remember, it is most important that your proposed pathway be chemically reasonable in other words, each hypothetical reaction that you propose should be very similar to a
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
319
Chapter 15: Oxidation and reduction
reaction pattern that we have seen in chapters 8-15. You should be able to put a name on
each step: for example, 'step 1 is a Claisen condensation; step 2 is a ketone reduction',
and so forth.
Also remember that there is usually no one correct way to approach problems like this they are puzzles to solve, and success will be dependent in large part on having
a solid grasp on the chemical 'tools' available to us: in other words, the biochemical
reaction types that we have been studying, starting with nucleophilic substitutions in
chapter 8.
P15.12: hypothetical 2-step transformations:
Each of the generalized transformations below would be expected to require two
enzymatic steps. Draw a reasonable pathway diagram for each transformation.
Example:
O
OP
R1
R2
NADH, H+
NAD+
R1
R2
ATP
ADP
Diagram:
O
R1
OH
R1
R2
OP
R1
R2
R2
The first step is a nicotinamide-dependent ketone reduction/hydrogenation, and step 2 is
ATP-dependent phosphorylation (ie. a kinase reaction). Note that the reducing agent in
the first step could also be NADPH)
a)
H
+
R1
H2N R2
N
R1
O
R2
b)
R1
SR2
R1
O
SR2
OH
O
c)
R1
R1
SR2
+
OH
320
O
R3SH
H3C
SR3
+
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
SR2
O
Chapter 15: Oxidation and reduction
d)
O
OH
d) R1
R1
R2
R2
e)
O
O
O
R1
+
CH3
R2
O
R1
H
R2
f)
O
O
R1
R1
R2
R2
P16.13: hypothetical 3-step transformations:
Each of the generalized transformations below would be expected to require three
enzymatic steps. Draw a reasonable pathway diagram for each transformation.
a)
O
O
R2
R1
+
O
R4
R3S
R1
H3N
R2
R4
b)
OH
R1
OH
R1
R2
OH
R2
OH
c)
O
O
O
R1
+
R1
R2
HO
R2
OH
OH
d)
R1
O
OH
+
R1
R2
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
O
R2
321
Chapter 15: Oxidation and reduction
P15.14: Now, let's try our hand at predicting the steps of some actual metabolic
pathways. For each, draw a complete pathway diagram. As needed, use any of the
coenzymes we have studied, water, and ammonia.
Note: you will probably find these quite challenging! Do not expect to be able to figure
them out in a few minutes - rather, think of them as puzzles to work on over a period of
time, sharing ideas and strategies with classmates. Remember, if one side of the
transformation is larger or more complex, start there and work towards the simpler
molecules. It is also a good idea, when applicable, to start the process by a) counting
carbons on each side of the transformation, and b) identifying the key bond being formed
(or broken) in the transformation.
a)
HO
HO
+ ???
H3CO
CO2
H3CO
vanillin
O
b)
HO
HO
H3CO
H3CO
H
N
O
O
+
O
O
c)
OH
OH
d)
O
3
322
H3C
CH3
O HO CH3
SCoA
O
OH
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
OPP
Chapter 15: Oxidation and reduction
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
323
Chapter 15: Oxidation and reduction
e)
H3CO
HO
NH2
HO
N
HO
CH3
O
+
HO
HO
f)
OP
O
+
O
CO2
O2C
O2C
succinate
OH
O
glyceraldehyde-3phosphate
P15.15: Propose a pathway diagram for each of the metabolic pathways below. Note that
some pathways contain steps that will be unfamiliar to you, and are therefore provided
already.
a) one cycle of fatty acid biosynthesis:
O
O
ACP
S
+
O
O
ACP
O
ACP
S
S
malonyl ACP
fatty acyl ACP
fatty acyl ACP
(2 carbons longer)
b) one cycle of fatty acid degradation:
O
SCoA
fatty acyl CoA
O
O
SCoA
fatty acyl CoA
(2 carbons shorter)
+
H3C
SCoA
acetyl CoA
c) Diabetics and people who adopt an extreme low-carbohydrate diet sometimes have
breath that smells like acetone, due to 'ketone body' formation that occurs when acetylCoA from fatty acid oxidation (see part (b) above) is not able to enter into the citric acid
324
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 15: Oxidation and reduction
cycle. Draw a pathway diagram showing how three molecules of excess acetyl CoA
combine to form acetone (all three acetyl-CoA molecules first link together, but one is
left over at the end of the process).
O
3 H3C
O
O
+
SCoA
H3C
acetyl CoA
H3C
CH3
SCoA
acetyl CoA
acetone
d) Pentose phosphate pathway (oxidative branch):
PO
HO
HO
OP
O
OH
OH
OH
OH
H
OH
O
ribulose-5-phosphate
glucose-6-phosphate
e) Citric acid (Krebs) cycle:
O
H3C
SCoA
acetyl-CoA
O2C
CO2
+
O
O
O2C
CO2
α-ketoglutarate
oxaloacetate
HSCoA
NAD+
(this reaction is analogous to pyruvate dehydrogenase,
which is discussed in section 17.x)
NADH/H+
CO2
CoAS
oxaloacetate
CO2
O
succinyl CoA
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
325
Chapter 15: Oxidation and reduction
f) proline biosynthesis:
H H
N
NH3
O
CO2
CO2
H
proline
glutamate
g) First half of lysine biosynthesis:
O
O
NH3
+
O
CO2
H3C
CO2
O2C
pyruvate
aspartate
N
CO2
tetrahydropicolinate
h) From the biosynthesis of membrane lipid in archaea: J. Bacteriol. 2003, 185, 1181
O
R
PPO
+
OH
R
PPO
OP
2x geranylgeranyl diphosphate dihydroxyacetone phosphate
(GGPP)
(DHAP)
NADH, H +, CTP, serine
NAD +, CMP, 3PPi
O 2C
NH 3
326
R
O
O O
P
O
O
O
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
R
Chapter 16: Free radical reactions
Chapter 16
Free radical reactions
Introduction
Imagine that you are an 18th century British sailor about set out with Commodore George
Anson to raid Spanish shipping fleets in the Pacific. You know full well that you are
signing up for a long and arduous ordeal, with months of constant seasickness, bad food,
cramped, unsanitary conditions, and brutal warfare. You are mentally ready for these
hardships, but what you are not prepared for is to watch your own body rot away – to
literally fall apart.
Below is a description of the suffering endured by many sailors of the time:
Some lost their very substance and their legs became swollen and puffed up while
the sinews contracted and turned coal-black and, in some cases, all blotched with
drops of purplish blood. Then the disease would creep up to the hips, thighs and
shoulders, arms and neck. And all the sick had their mouths so tainted and their
gums so decayed that the flesh peeled off down to the roots of their teeth, which
almost all fell out. . .
There were devastating neurological as well as physiological effects. Scurvy had the
ability to inhibit a person's normal restraints on emotion: they became intensely homesick
and nostalgic, wept at the slightest disappointment, and screamed in agony upon smelling
the scent of flower blossoms drifting across the water from a nearby shore.
The disease afflicting the sailors was scurvy, which we now know is caused by a
deficiency of vitamin C in the diet. European sea voyagers in the 18th century and earlier
subsisted mainly on a diet of salted meat, hard biscuits, pea soup, oatmeal, and beer.
After the first couple of weeks at sea, fresh fruits and vegetables - and the nutrients they
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
327
Chapter 16: Free radical reactions
contained – were all consumed or spoiled. The salted meat and hardtack diet provided
salt and calories, but little else of nutritional value.
Although it is rare now, scurvy has plagued sailors for centuries, with records of its
occurrence on ships going back as far as the 15th century voyages of Magellan and Vasco
de Gama, both of whom lost up to three quarters of their crew to the disease on long
ocean crossings. Various cultures made the connection between scurvy and diet, and
learned effective preventative measures: sailors with the 16th century French explorer
Jacques Cartier, for example, were cured of their scurvy upon arriving in Canada and
taking the advice of native people to eat the leaves and bark of pine trees. These were
lessons, unfortunately, that often had to be relearned time and again, as the knowledge
gained by one culture was not effectively recorded and passed along to others.
Vitamin C, or ascorbic acid as it is known to chemists, plays an essential helping role in a
variety of essential biochemical reactions. Most living things are able to synthesize
ascorbic acid – the exceptions include humans and other higher primates, several species
of bats, and some rodents such as guinea pigs and capybaras. Humans lack the last
enzyme in the ascorbic acid biosynthetic pathway, L-gulonolactone oxidase. (EC 1.1.3.8)
(You were invited to propose the mechanism for this redox reaction in problem 15.10).
HO
O
O
O2
HO
H2O2
HO
O
O
HO
keto
HO
OH
(FADH2]
O
OH
L-gulonolactone
spontaneous tautomerization
HO
O
O
HO
enol
HO
OH
ascorbic acid
fig 1a
Because we cannot make our own ascorbic acid, we need to get it in our diet. It is
abundant in many plant-based foods, citrus fruits in particular. The traditional diet of the
Inuit people of the arctic region contains virtually no plant products, but vitamin C is
obtained from foods such as kelp, caribou livers, and whale skin. For a time in the 18th
century, the observation that citrus fruits quickly cured scurvy led to the practice of
including in a ship's stores a paste prepared from boiled lemon juice. Unfortunately,
ascorbic acid did not survive the boiling process, rendering the paste ineffective against
328
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
scurvy. Captain James Cook, the legendary explorer and the first European to make it to
the east coast of Australia and the Hawaiian islands, brought along sourkraut (fermented
cabbage), a somewhat more effective vitamin C supplement. According to his own
account, Cook's sailors at first refused to eat the pungent preparation, so the captain
engaged in a little psychological trickery: he declared that it would only be served to
officers. The enlisted sailors quickly took offense, and demanded their own sourkraut
ration.
***
The biochemical role of ascorbic acid is to facilitate the transfer of single electrons in a
variety of redox reactions - note here the emphasis on single electrons, as opposed to the
redox reactions we studied in chapter 15 in which electrons were transferred in pairs. The
subject of this chapter is single-electron chemistry, and the free radical intermediates that
are involved in single electron reaction steps.
Later in this chapter we will learn the chemical details of why ascorbic acid deficiency
causes scurvy, how the act of breathing makes you get old, how polystyrene packing
foam is made, and other interesting applications of single-electron chemistry. But first
we need to cover some basics ideas about single electron chemical steps, and the free
radical intermediates that result from them.
Section 16.1: Overview of single-electron reactions and free radicals
Beginning with acid-base reactions in chapter x and continuing though the chapters on
nucleophilic substitution, carbonyl addition, acyl substitution, α-carbon chemistry, and
electrophilic reactions , we have been studying reaction mechanisms in which both
electrons in a covalent bond or lone pair move in the same direction. In this chapter, we
learn about reactions in which the key steps involve the movement of single electrons.
Single electron movement is depicted by single-barbed 'fish-hook' arrows (as opposed to
the familiar double-barbed arrows that we have been using throughout the book to show
two-electron movement).
curved, two-barbed arrow:
two-electron movement
curved, single-barbed arrow:
single-electron movement
fig 1
Single-electron mechanisms involve the formation and subsequent reaction of free
radical species, highly unstable intermediates that contain an unpaired electron. Free
radicals are often formed from homolytic cleavage, an event in which the two electrons
in a breaking covalent bond move in opposite directions. The bond in molecular chlorine,
for example, is subject to homolytic cleavage when chlorine is subjected to heat or light.
The result is two chlorine radicals. Note that each radical has a formal charge of zero.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
329
Chapter 16: Free radical reactions
single-barbed arrows
chlorine radicals
Cl
Cl
+
Cl
Cl
homolytic cleavage
(two electrons in breaking bond go with different atoms)
fig 2
In contrast, essentially all of the reactions we have studied up to now involve bondbreaking events in which both electrons move in the same direction: this is called
heterolytic cleavage.
double-barbed arrow
CH3
CH3
H3C C
Cl
H3C
H3C
C
+
Cl
CH3
heterolytic cleavage
(both electrons in breaking bond go with the same atom)
fig 3
Two other homolytic cleavage reactions that we will see in this chapter can be described
as 'radical hydrogen atom abstraction' and 'radical alkene addition':
hydrogen abstraction by a radical
H H
H C C
H
H H
Cl
+
H C C
H H
H Cl
H H
alkene addition by a radical
Cl
H
H
H
R
R
C C
R
Cl
C C
H
H
fig 4
Single-electron reaction mechanisms involve the formation of radical species, and in
organic reactions these are often carbon radicals. A carbon radical is sp2 hybridized, with
three σ bonds arranged in trigonal planar geometry and the single unpaired electron
occupying an unhybridized p orbital. Contrast this picture with a carbocation reactive
330
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
intermediate, which is also sp2 hybridized with trigonal planar geometry but with an
empty p orbital.
R
p orbital contains a
single unpaired electron R
RR
RR
carbocation
carbon radical
(neutral)
empty p orbital
fig 5
When we studied electrophilic reactions in chapter 14, a major concern when evaluating
possible mechanisms was the stability of any carbocation intermediate(s). Likewise, the
stability of proposed radical intermediates is of great importance when evaluating the
likelihood of possible single-electron mechanisms. Fortunately, the trend in the stability
of carbon radicals parallels that of carbocations (section 8.5): tertiary radicals, for
example, are more stable than secondary radicals, followed by primary and methyl
radicals. This should make intuitive sense, because radicals, like carbocations, are
electron deficient, and thus are stabilized by the electron-donating effects of nearby alkyl
groups.
R
H
H
H
R C
R C
H C
H C
H
R
R
R
3o radical
2o radical
1o radical
most stable
methyl radical
least stable
fig 6
Benzylic and allylic radicals are more stable than alkyl radicals due to resonance effects an unpaired electron (just like a positive or negative charge) can be delocalized over a
system of conjugated π bonds. An allylic radical, for example, can be pictured as a
system of three parallel p orbitals sharing three electrons.
fig 7
The drawing below shows how a benzylic radical is delocalized to three additional
carbons around the aromatic ring:
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
331
Chapter 16: Free radical reactions
fig 8
Exercise 16.1: Just as phenolate ions are less reactive (less basic) than alkoxide ions,
phenolic radicals are less reactive than alkoxide radicals. Draw one resonance
contributor of a phenolic radical showing how the radical electron is delocalized to a ring
carbon. Include electron-movement arrows.
O
phenolic radical
While radical species are almost always very reactive and short-lived, in some extreme
cases they can be unreactive. One example of an inert organic radical structure is shown
below.
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
C
Cl
Cl
Cl Cl
Cl
Cl
Cl
fig 10
The already extensive benzylic resonance stabilization is further enhanced by the fact
that the large electron clouds on the chlorine atoms shield the radical center from external
reagents. The radical is, in some sense, inside a protective 'cage'.
332
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
Exercise 16.2: Draw a resonance contributor of the structure above in which the unpaired
electron is formally located on a chlorine atom (include electron movement arrows)
Section 16.2: Radical chain reactions
Because of their high reactivity, free radicals have the potential to be extremely powerful
chemical tools - but as we will see in this chapter, they can also be extremely harmful in a
biological/environmental context. Key to understanding many types of radical reactions
is the idea of a radical chain reaction.
Radical chain reactions have three distinct phases: initiation, propagation, and
termination. We'll use a well-known example, the halogenation of an alkane such as
ethane, to illustrate. The overall reaction is:
CH3CH3 + Cl2
hν or Δ
CH3CH2Cl + HCl
(hν means light
Δ means heat)
fig 11
The initiation phase in a radical chain reaction involves the homolytic cleavage of a
weak single bond in a non-radical compound, resulting in two radical species as products.
Often, heat or light provides the energy necessary to overcome an energy barrier for this
type of event. The initiation step in alkane halogenation is homolysis of molecular
chlorine (Cl2) into two chlorine radicals. Keep in mind that that virtually all radical
species, chlorine radicals included, are highly reactive.
Cl
Cl
2 Cl
fig12
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
333
Chapter 16: Free radical reactions
The propagation phase is the 'chain' part of chain reactions.
a)
H
H
H
H3C C
+
H3C C
Cl
HCl
H
H
b)
H
H
Cl
H3C C
Cl
H3C C Cl
+
Cl
H
H
(back to step (a) to react with
another ethane molecule)
fig 13
Once a reactive free radical (chlorine radical in our example) is generated in the initiation
phase, it will react with relatively stable, non-radical compounds to form a new radical
species. In ethane halogenation, a chlorine radical generated in the initiation step first
reacts with ethane in a hydrogen abstraction step, generating HCl and an ethyl radical
(part a above). Then, the ethyl radical reacts with another (non-radical) Cl2 molecule,
forming the chloroethane product and regenerating a chlorine radical (part b above).
This process repeats itself again and again, as chlorine radicals formed in part (b) react
with additional ethane molecules as in part (a).
The termination phase is a radical combination step, where two radical species happen
to collide and react with each other to form a non-radical product and 'break the chain'.
In our ethane chlorination example, one possible termination event is the reaction of a
chlorine radical with an ethyl radical to form chloroethane.
H
H3C C
H
H
Cl
H3C C Cl
H
fig 14
Exercise 16.3: Draw two alternative chain termination steps in the ethane chlorination
chain reaction. Which one leads to an undesired product?
Because radical species are so reactive and short-lived, their concentration in the reaction
mixture at any given time is very low compared to the non-radical components such as
ethane and Cl2. Thus, many cycles of the chain typically occur before a termination event
334
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
takes place. In other words, a single initiation event leads to the formation of many
product molecules.
Compounds which readily undergo homolytic cleavage to generate radicals are called
radical initiators. As we have just seen, molecular chlorine and bromine will readily
undergo homolytic cleavage to form radicals when subjected to heat or light. Other
commonly used as radical initiators are peroxides and N-bromosuccinimide (NBS).
Cl
Cl
RO OR
2 Cl
hν or Δ
hν or Δ
2 RO
peroxide
O
O
Br
N
N
hν or Δ
O
+
Br
O
NBS
fig 15
Section 16.3: Useful polymers formed by radical chain reactions
Many familiar household materials polymers made from radical chain reaction processes.
Polyethylene (PET), the plastic material used to make soft drink bottles and many other
kinds of packaging, is produced by the radical polymerization of ethylene (ethylene is a
common name for what we call 'ethene' in IUPAC nomenclature). The process begins
when a radical initiator such as benzoyl peroxide undergoes homolytic cleavage at high
temperature:
initiation
O
O
O
O
2
Δ
O
benzoyl peroxide
O
benzoyl radical
fig 16
In the propagation phase, the benzoyl radical (X• in the figure below) adds to the double
bond of ethylene, generating a new organic radical.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
335
Chapter 16: Free radical reactions
propagation
O
X
X
H
a)
H C C H
H
H
X
C C H
H
H
H H H H
H
b1)
X
C C H
H
H
H C C H
H
H
X
C C C C
H H H H
H H H H
b2)
O
=
X
H H H H H H
C C C C
H C C H
H
H
H H H H
X
C C C C C C
H H H H H H
etcetera
fig 17
Successive ethylene molecules add to the growing polymer, until termination occurs
when two radicals happen to collide.
termination
X
n
X
X
X
n
this notation indicates that there are many
repeating -CH2CH2- monomers in the chain
fig 18
The length of the polymer is governed by how long the propagation phase continues
before termination, and can usually be controlled by adjusting reaction conditions.
336
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
Other small substituted alkene monomers polymerize in a similar fashion to form familiar
polymer materials. Two examples are given below.
X
styrene
X
n
Ph
Ph
polystyrene
Ph
(packing foam)
X
Cl
vinyl chloride
X
n
Cl
Cl
Cl
polyvinyl chloride (PVC)
(household plumbing, insulation)
fig 19
Exercise 16.4: Show a mechanism for the formation of a 2-unit long section of
polystyrene, starting with the monomer and benzoyl peroxide initiator. Keep in mind the
relative stability of different radical intermediates.
A common way to separate proteins in the biochemistry lab is through a technique called
polyacrylamide gel electrophoresis (PAGE). The polyacrylamide gel is formed through
radical polymerization of acrylamide monomer, with the ammonium salt of persulfate
used as the radical initiator.
n
O
NH2
acrylamide
O
=
X
NH2
X
X
X
(where X = CONH2)
polyacrylamide
O
X
O
O S O O S O
O
O
persulfate
(radical initiator)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
337
Chapter 16: Free radical reactions
fig 20
In the end of chapter problems, you will be invited to propose a mechanism showing how
a molecule called 'bis-acrylamide' serves as a 'crosslinker' between linear polyacrylamide
chains to allow for formation of a net-like structure for the PAGE gel.
Section 16.4: Destruction of the ozone layer by a radical chain reaction
The high reactivity of free radicals and the multiplicative nature of radical chain reactions
can be useful in the synthesis of materials such as polyethylene plastic - but these same
factors can also result in dangerous consequences in a biological or ecological context.
You are probably aware of the danger posed to the earth's protective stratospheric ozone
layer by the use of chlorofluorocarbons (CFCs) as refrigerants and propellants in aerosol
spray cans. Freon-11, or CFCl3, is a typical CFC that was widely used until late in the
20th century. It can take months or years for a CFC molecule to drift up into the
stratosphere from the surface of the earth, and of course the concentration of CFCs at this
altitude is very low. Ozone, on the other hand, is continually being formed in the
stratosphere. Why all the concern, then, about destruction of the ozone layer - how could
such a small amount of CFCs possibly do significant damage? The problem lies in the
fact that the process by which ozone is destroyed is a chain reaction, so that a single CFC
molecule can initiate the destruction of many ozone molecules before a chain termination
event occurs.
Although there are several different processes by which the ozone destruction process
might occur, the most important is believed to be the chain reaction shown below.
F
Cl C
Cl
Cl
O O O
Cl
Cl
O
O Cl
Cl
O O Cl
Cl
O O
F
step 1
step 2
step 3
step 4
step 5
C +
Cl
Cl Cl
+ O2
Cl
O
Cl
O O Cl
Cl
O O + Cl (back to step 2)
O O + Cl (back to step 2)
O2
fig 21
First, a CFC molecule undergoes homolytic cleavage upon exposure to UV radiation,
resulting in the formation of two radicals (step 1). The chlorine radical rapidly reacts
338
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
with ozone (step 2) to form molecular oxygen and a chlorine monoxide radical. Step 3
appears to be a chain termination step, as two chlorine monoxide radicals combine. The
Cl2O2 condensation product, however, is highly reactive and undergoes two successive
homolytic cleavage events (steps 4 and 5) to form O2 and two chlorine radicals, which
propagates the chain.
To address the problem of ozone destruction, materials chemists have developed new
hydrofluorocarbon refrigerant compounds. The newer compounds contain carbonhydrogen bonds, which are weaker than the carbon-halogen bonds in CFCs , and thus are
susceptible to homolytic cleavage caused by small amounts of hydroxide radical present
in the lower atmosphere:
CF3
CF3
F C
H H
C
OH
+ H2O
FH
fig 22
This degradation occurs before the refrigerant molecules have a chance to drift higher up
to the stratosphere where the ozone plays its important protective role. The degradation
products are quite unstable and quickly degrade further into relatively harmless byproducts. The hydroxide radical is sometimes referred to as an atmospheric 'detergent'
due to its ability to degrade escaped refrigerants and other volatile organic pollutants.
Hydrofluorocarbons do, however, act as greenhouse gases, and are thought to contribute
to climate change.
Section 16.5: Oxidative damage, vitamin C, and scurvy
While the hydroxide radical can be a beneficial 'detergent' in the atmosphere, it is harmful
when present in a living cell. Hydroxide radical is one of the reactive oxygen species
(ROS) that we learned about in chapter 15. Recall that ROS are continuously produced
as minor but harmful side-products in the reduction of O2 to H2O in respiration.
You may recall from your general chemistry course that molecular oxygen exists in two
states: 'singlet' oxygen has a double bond and no unpaired electrons, while 'triplet' oxygen
has a single O-O bond and two unpaired electrons. Molecular orbital theory - and
experimental evidence - show that the triplet state is lower in energy.
O O
O O
triplet oxygen
(ground state)
singlet oxygen
(excited state)
fig 23
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
339
Chapter 16: Free radical reactions
ROS are highly reactive oxidizing agents, capable of inflicting damage to DNA, proteins,
and the lipids of cell membranes - they are thought to play a major role in the aging
process. Hydroxide radical, for example, can initiate a radical chain reaction with the
hydrocarbon chain of an unsaturated membrane lipid molecule, resulting in the
formation of lipid hydroperoxide.
R
C
R
step 1
R
C
R
step 2
C
H O
R
H
H H
O
O O
(triplet) oxygen
OH
section of lipid chain
R
R
H H
C
R
another lipid
molecule
step 3
H
R
C
R
+
R
C
H O
R
OH
lipid hydroperoxide
(damaged)
lipid radical
(continues chain reaction)
fig 24
The allylic lipid radical formed as the result of homolytic hydrogen abstraction by
hydroxide radical (step 1 above) reacts with one of the unpaired electrons in triplet
oxygen (step 2) forming a peroxy radical. This radical species in turn homolytically
abstracts a hydrogen from another lipid molecule (step 3), thus propagating the chain.
Many edible plants contain various antioxidant compounds, also known as 'free radical
scavengers', which serve to protect cells from the oxidative effects of hydroxide radical
and other harmful radical intermediates. Simply put, a free radical scavenger is a
molecule that reacts with a potentially damaging free radical species, forming a more
stable radical species which can be metabolized by the body before any damage is done
to cell constituents.
radical
scavenger
X
harmful
free radical
340
more stable
radical
X +
+ Y
non-radical
Y
metabolized, excreted
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
In the introduction to this chapter, we learned about scurvy, the disease long dreaded by
sailors, and how it is caused by a deficiency of ascorbic acid (vitamin C) in the diet. We
will soon get to the connection between ascorbic acid and scurvy, but first, let's look at
how ascorbic acid functions as a free radical scavenger in your body.
The pKa of ascorbic acid is about 4.1, so in a physiological environment it exists mainly
as ascorbate anion, the conjugate base. When ascorbate encounters a hydroxide radical
(or any other potentially damaging radical species), it donates a single electron, thus
reducing the hydroxide radical to hydroxide ion and becoming itself an ascorbyl radical.
HO
O
O
R
O
O
HO
O
O
OH
ascorbate
O
R
O
OH
ascorbyl radical
OH
hydroxyl radical
O
OH
+
OH
hydroxide ion
fig 24a
The ascorbyl radical is stabilized by resonance. The end result of this first step is that a
very reactive, potentially harmful hydroxide radical has been 'quenched' to hydroxide ion
and replaced by a much less reactive (and thus less harmful) ascorbyl radical.
The ascorbyl radical can then donate a second electron to quench a second hydroxide
radical, resulting in the formation of dehydroascorbate, the oxidized form of ascorbate.
O
R
HO
O
O
HO
O
+
:B
O
OH
O H
O
OH
O
dehydroascorbate
fig 24b
One ascorbate molecule is thus potentially able to scavenge two harmful radical species.
Dehydroascorbate is subsequently either broken down and excreted, or else
enzymatically recycled (reduced) back to ascorbic acid. You were invited to propose a
mechanism for the latter (redox) step in problem 15.10. J. Am. Coll. Nutr., 2003, 22, 18
We learned in the introduction to this chapter about the gruesome effects of long-term
ascorbic acid deficiency. What, then, is the chemical connection between ascorbic acid
and scurvy?
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
341
Chapter 16: Free radical reactions
The symptoms association with scurvy are caused by the body's failure to properly
synthesize collagen, the primary structural protein in our connective tissues. Essential to
the stability of collagen is its ability to form a unique triple-helical structure, in which
three protein strands coil around each other like a woven rope. Collagen strands are not
able to pack together properly into their triple helix structure unless certain of their
proline amino acid residues are hydroxylated: the electronegative OH group on
hydroxyproline causes the five-membered ring in the amino acid to favor a particular
'envelope' conformation (section 3.2) as well as the 'trans' peptide conformation, both of
which are necessary for stable triple-helix formation.
HO
protein
N
protein
N
O
protein
proline residue
O
protein
hydroxyproline residue
OH
protein
protein
N
N
OH
O
O
protein
protein
favored envelope conformation:
allows for triple-helix formation
HO
HO
protein
protein
N
O
N
O
protein
protein
cis peptide conformation
O
O
trans peptide conformation:
allows for triple-helix formation
fig 26a
Proline hydroxylase, the enzyme responsible for this key modification reaction, depends
in turn upon the presence of ascorbate. The hydroxylating reaction is complex, and
involves electron-transfer steps with enzyme-bound iron - mechanistic details that are
well outside of our scope here, but which you may learn about in a bioinorganic
342
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
chemistry course. It is enough for us to know that iron starts out in the Fe+2 state, and
during the course of the reaction it loses an electron to assume the Fe+3 state. In order for
the enzyme to catalyze another reaction, the iron must be reduced back to its active Fe+2
state - it must accept a single electron. The donor of this single electron is ascorbate.
(For more information, see Crit. Rev. Biochem. Mol. Biol. 2010, 405, 106.)
So, to sum up: If we fail to get enough ascorbic acid in our diet (in other words, if we
don't eat our fruits and vegetables!) the iron in our proline hydroxylase enzymes won't be
returned to the active Fe+2 state, so the catalytic cycle is broken and we can't turn prolines
into hydroxyprolines. Without the hydroxy group, the proline residues of our collagen
proteins won't assume the proper conformation, and as a consequence the collagen triple
helix structures will be unstable. At physiological temperature, our collagen will literally
melt apart - and with it, our gums, our capillaries, and anything else held together by
connective tissue. This is scurvy.
You have probably heard that many fruits and vegetables contain natural 'polyphenol'
antioxidant compounds that are thought to be beneficial to our health. Apigenin, for
example, is found in parsley and celery, while the skins of grapes used to produce red
wine are particularly rich in resveratrol, as well as many other polyphenols. Curcumin is
the compound responsible for the distinctive yellow color of turmeric, a ubiquitous spice
in Indian cuisine.
OH
HO
OH
O
OH
HO
O
OH
apigenin
resveratrol
HO
OH
O
O
CH3
O
OH
CH3
curcumin
fig 25
While much remains to be learned about exactly how these polyphenols exert their
antioxidant effects, it is likely that they, like ascorbic acid, act as radical scavengers. For
example, resveratrol could donate a single electron (and a proton) to hydroxide radical to
reduce it to water. The phenolic radical that results is stabilized by resonance, and is
much less likely than hydroxide radical to cause damage to important biomolecules in the
cell.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
343
Chapter 16: Free radical reactions
radical stabilized by resonance
O
H
O
OH
HO
HO
OH
OH
+ H2O
fig 26
Section 16.6: Flavin as a a single-electron carrier
In chapter 15 we saw how a nicotinamide and flavin coenzymes can act as acceptors or
donors of two electrons in hydride-transfer redox steps. Recall that it was mentioned that
flavin, (but not nicotinamide) can also participate in single-electron transfer steps through
a stabilized radical intermediate called a semiquinone.. Frey p. 162 fig 3-30; Silverman
p. 122 sch. 3.34; J Phys Chem A. 2013, 117, 11136 fig 2)
1-electron acceptor
H
N
1-electron acceptor
H
O
N
N
N
R
H
reduced flavin
(FADH2 or FMNH2)
H
N
O
N
:B
O
O
N
N
H
N
O
N
R
flavin semiquinone
(radical)
N
N
H
O
R
oxidized flavin
(FAD or FMN)
fig 27
Note in this reaction that overall, flavin loses or gains two electrons and two protons, just
like in the flavin-dependent redox reactions we saw in chapter 15. The difference here is
that the electrons are transferred one at a time, rather than paired in the form of a hydride
ion.
Two important examples single-electron acceptor species in human metabolism are
ubiquinone (coenzyme Q) and the oxidized form of cytochrome. Ubiquinone is a
coenzyme that can transfer single electrons via a semiquinone state analogous to that of
flavin, and cytochrome is a protein containing a 'heme' iron center which shuttles
between the Fe+3 (oxidized) and Fe+2 (reduced) state.
344
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
H+
e-
O
H+
e-
OH
OH
H3CO
CH3
H3CO
CH3
H3CO
CH3
H3CO
R
H3CO
R
H3CO
R
O
O
OH
oxidized ubiquinone
(coenzyme Q)
ubiquinone semiquinone
(radical)
reduced ubiquinone
(QH2)
e-
Fe+3
(heme center in cytochrome protein)
Fe+2
fig 27a
Further discussion of the mechanisms of single-electron flavin reactions is beyond our
scope here, but when you study the 'respiratory chain' in a biochemistry course you will
gain a deeper appreciation for the importance of flavin in single-electron transfer
processes.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
345
Chapter 16: Free radical reactions
Problems
P16.1: Plexiglass is a polymer of methyl methacrylate. Show a mechanism for the first
two propagation steps of polymerization (use X• to denote the radical initiator), and show
a structure for the plexiglass polymer. Assume an alkene addition process similar to that
shown in the text for polyethylene.
OCH3
O
methyl methacrylate
P16.2: In section 16.3 we saw how acrylamide polymerizes to form the polyacrylamide
used in PAGE protein gels. Polyacrylamide by itself is not sufficient by itself to form the
gel - the long polyacrylamide chains simply slip against each other, like boiled spaghetti.
To make a PAGE gel, with pores for the proteins to slip through, we need a crosslinker something to tie the chains together, forming a three-dimensional web-like structure.
Usually, a small amount of bis-acrylamide is added to the acrylamide in the
polymerization mixture for this purpose.
O
O
N
N
H
H
bis-acrylamide
Propose a radical mechanism showing how bis- acrylamide might form crosslinks
between two polyacrylamide chains.
P16.3: Resveratrol is a natural antioxidant found in red wine (see section 16.5).
a) Draw one resonance structure to illustrate how the resveratrol radical shown in figure
26 is delocalized by resonance.
b) Indicate all of the carbons on your structure to which the radical can be delocalized.
c) Draw an alternate resveratrol radical (one in which a hydrogen atom from one of the
other two phenolic groups has been abstracted). To how many carbons can this radical be
delocalized?
346
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
d) Draw two resonance contributors of a curcumin radical, one in which the unpaired
electron is on a phenolic oxygen, and one in which the unpaired electron is on a ketone
oxygen.
P16.4: Draw the radical intermediate species that you would expect to form when each of
the compounds below reacts with a radical initiator.
a)
b)
OH
c)
d) HO
OH
SH
P16.5: Azobis(isobutyronitrile) is a widely used radical initiator which rapidly undergoes
homolytic decomposition when heated.
CH3
H3C C
CN
CH3
N N C CH3
CN
Azobis(isobutyronitrile)
Predict the products of this decomposition reaction, and show a likely mechanism. What
is the thermodynamic driving force for homolytic cleavage?
P16.6:
a) When 2-methylbutane is subjected to chlorine gas and heat, a number of isomeric
chloroalkanes with the formula C5H11Cl can form. Draw structures for these isomers, and
for each draw the alkyl radical intermediate that led to its formation.
b) In part a), which is the most stable radical intermediate?
c) In the reaction in part a), the relative abundance of different isomers in the product is
not exclusively a reflection of the relative stability of radical intermediates. Explain.
P16.7: We learned in chapter 14 that HBr will react with alkenes in electrophilic addition
reactions with 'Markovnikov' regioselectivity. However, when the starting alkene
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
347
Chapter 16: Free radical reactions
contains even a small amount of contaminating peroxide (which happens when it is
allowed to come into contact with air), a significant amount of 'anti-Markovnikov'
product is often observed.
a) Propose a mechanism for formation of the anti-Markovnikov addition product when 1butene reacts with HBr containing a small amount of benzoyl peroxide
b) Predict the product and propose a mechanism for the addition of ethanethiol to 1butene in the presence of peroxide.
P16.8: In section 11.5 we learned that aspirin works by blocking the action of an enzyme
that catalyzes a key step in the biosynthesis of prostaglandins, a class of biochemical
signaling molecules. The enzyme in question, prostaglandin H synthase (EC 1.14.99.1)
catalyzes the reaction via several single-electron steps. First, an iron-bound oxygen
radical in the enzyme abstracts a hydrogen atom from arachidonate. The arachidonate
radical intermediate then reacts with molecular oxygen to form a five-membered oxygencontaining ring, followed by closure of a cyclopentane ring to yield yet another radical
intermediate. (Biochemistry 2002, 41, 15451.)
CO2
O2
CO2
O
O
O
arachidonate
enz
prostaglandin H2
CO2
O
O
(show mechanims to this point)
Propose a mechanism for the steps of the reaction that are shown in this figure.
P16.9: Some redox enzymes use copper to assist in electron transfer steps. One
important example is dopamine β-monooxygenase (EC 1.14.1.1), which catalyzes the
following reaction:
348
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 16: Free radical reactions
OH
HO
HO
O2
NH2
+ H2O
NH2
HO
HO
norepinephrine
dopamine
The following intermediates have been proposed:
(Biochemistry 1994, 33, 226), Silverman p. 222)
enz
enz
enz
enz
Cu
enz
Cu
step 1
step 2
O
O
OH
R
enz
OH
NH2
OH
H2O
O
NH2
R
Cu
NH2
R
dopamine
step 3
enz
enz
enz
enz
Cu
enz
Cu
enz
OH
Cu
step 5
OH
R
O
OH
O
step 4
O
NH2
R
NH2
R
OH
NH2
Draw mechanistic arrows for steps 1-4.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
349
Chapter 17: Coenzymes
Chapter 17
Coenzymes
The old black and white photograph is haunting. A young boy, perhaps 10 or 11 years
old, huddles against a wall outside a soup kitchen, his mouth in an odd twisted shape that
could be expressing either pain or defiance, his eyes staring straight into those of the
viewer. Tucked into his pants, almost like a pistol in a holster, is a metal spoon.
The photograph was taken in the Netherlands in 1945, at the height of what the Dutch
people still refer to simply as "The Hunger Winter". With the western part of the country
still occupied by the Nazis, the Dutch resistance government, based in London, had called
for a railway strike with the aim of stopping German troop movements before a planned
airborne invasion by Allied forces. In retaliation, the Germans cut off all food shipments
to cities in the western Netherlands. The Allied invasion failed to liberate the country,
and the winter of 1944-1945 turned out to be bitterly cold. With food supplies dwindling,
rations were cut first to 1000 calories per day, then to 500. People resorted to eating grass
and tulip bulbs just to stay alive. Over 20,000 people died of starvation before food
shipments were restored in the spring.
As tragic as the Hunger Winter was for the Dutch people, some good did come from it.
For medical scientists, the event became a unique 'social experiment': unlike most
episodes of famine throughout human history, the Hunger Winter had a clearly defined
beginning, end, and geographic boundary, and it occurred in a technologically advanced
society that continued to keep thorough records before, during, and after the ordeal.
Scientists knew exactly who suffered from famine and for precisely how long, and in the
years that followed they were able to look at the long-term effects of famine, particularly
on developing embryos. Researchers found that babies who had been conceived during
the famine were born with a significantly higher incidence of neurological birth defects
350
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
such as spina bifida, a condition in which a portion of the neural tube protrudes from
between vertebrae which did not properly fuse together during fetal development. Later
in life, members of this same cohort of 'famine babies' were more likely to be obese, and
to suffer from schizophrenia.
These initial findings spurred interest in further research into the consequences of
prenatal deprivation. In particular, carefully controlled studies later led to the recognition
of the importance of the vitamin folate in ensuring proper neurological development in
early-term fetuses.
H2N
H
N
N
N
N
O
H
N
H
N
folate
O
CO2
CO2
fig 1a
Folate - the conjugate base of folic acid - is an organic coenzyme: a helper molecule that
binds in the active site of certain enzymes and plays a critical role in the biochemical
reaction being catalyzed. Recall that we have seen coenzymes at work before: SAM,
ATP, NAD(P)+ and NAD(P)H, flavin and glutathione are all important coenzymes with
which we are already familiar.
Because prenatal folate deficiency was found to be directly related to the incidence of
spina bifida and other neural tube defects, health officials in the United States and many
other countries changed their official guidelines to include a specific recommendation
that women begin taking folate supplements as soon as they knew that they are pregnant,
or better yet as soon as they begin trying to become pregnant. A number of studies
conducted during the 1980s and early 1990s consistently showed that folate
supplementation correlated with a 50-70% reduction in neural tube defects.
The molecular role of folate in prenatal neurological development is not understood in
detail, but most researchers agree that it probably has a lot to do with DNA biosynthesis.
Like S-adenosyl methionine (SAM), folate functions in 1-carbon transfer reactions,
including several critical steps in the nucleic acid biosynthesis pathways. The rapidly
dividing cells of the developing brain of an early term fetus appear to be especially
sensitive to folate deficiency in the mother's diet: insufficient folate leads to impaired
DNA biosynthesis, which in turn leads to defects in brain development.
Folate also serves as a 1-carbon donor in the pathway by which SAM is regenerated after
it donates a methyl group. You may recall from the introduction to chapter 8 that
methylation of cytosine bases in DNA by SAM results in permanent changes to a
individual's genome - this was the reason why the two 'identical' twin sisters in that
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
351
Chapter 17: Coenzymes
introductory story turned out to be, as they grew older, not so identical after all. It is
likely that the folate deprivation that afflicted expectant mothers during the Dutch Hunger
Winter also caused epigenetic changes (in other words, changes in the extent of DNA
methylation) in their developing fetuses, which decades later manifested in the form of an
increased incidence of conditions such as obesity and schizophrenia. All the more
reason, we now know, to make sure that women get plenty of folate in their diet early in
the first trimester of pregnancy.
In this final chapter, we focus on the organic chemistry of folate, along with three other
coenzymes: pyridoxal phosphate, thiamine diphosphate, and lipoamide. Humans can
synthesize lipoamide, but we depend on dietary sources for the other three: pyridoxal
phosphate is a form of vitamin B6, and thiamine diphosphate is a form of vitamin B1. In
a mechanistic sense, there is really nothing new in this chapter. All of the reaction
mechanism types that we will see are already familiar to us, ranging from nucleophilic
substitutions (chapter 8) to disulfide exchanges (chapter 15). We will soon see, however,
how each coenzyme plays its own specific and crucial role in assisting enzymes with the
catalysis of key reactions of metabolism. We will begin with pyridoxal phosphate and its
various roles in amino acid metabolism.
Additional reading:
http://www.naturalhistorymag.com/features/142195/beyond-dna-epigenetics
Section 17.1: Pyridoxal phosphate (Vitamin B6)
The coenzyme pyridoxal phosphate (commonly abbreviated PLP) is the active form of
vitamin B6, or pyridoxine.
H
OH
OH
HO
N
CH3
O
OH
PO
N
CH3
H
H
pyridoxine (vitamin B6)
pyridoxal phosphate (PLP)
fig 1
PLP is required for over 100 different reactions in human metabolism, primarily in the
various amino acid biosynthetic and degradation pathways. The essential function of PLP
is to act as an 'electron sink', stabilizing a negative formal charge that develops on key
reaction intermediates. Some of reactions will be familiar to you from chapter 12 and 13:
we will see examples, for instance, of PLP-dependent α-carbon racemization, as well as
aldol- and Claisen-type reactions. Other reactions will be less familiar: for example, the
352
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
participation of PLP allows for decarboxylation of amino acids, a chemical step which
would be highly unlikely without the coenzyme, and PLP is also required for a very
important class of biochemical transformation called 'transamination', in which the amino
group of an amino acid is transferred to an acceptor molecule. Before we dive into the
reactions themselves, though, we need to begin by looking at a key preliminary step that
is common to all of the PLP reactions we will see in this section.
17.1A: PLP in the active site: the imine linkage
The common catalytic cycle of a PLP-dependent enzyme begins and ends with the
coenzyme covalently linked to the enzyme's active site through an imine linkage between
the aldehyde carbon of PLP and the amine group of a lysine residue (see section 10.5 to
review the mechanism for imine formation). For a PLP-dependent enzyme to become
active, a PLP molecule must first enter the active site of an enzyme and form an imine
link to the lysine. This state is often referred to as an external aldimine.
enz
side chain of active-site
lysine residue
NH 2
enz
imine linkage
H
O
H
H 2O
N
OH
PO
OH
PO
N
CH3
N
H
free PLP
CH3
H
PLP linked to enzymatic lysine
(external aldimine)
fig 2
The first step of virtually all PLP-dependent reactions is transimination (section 10.5), as
the amino group on the amino acid substrate displaces the amino group of the enzymatic
lysine. This state - where the coenzyme is covalently linked to the substrate or product of
the reaction - is often referred to as an internal aldimine.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
353
Chapter 17: Coenzymes
O
R
O
NH 3
enz
R
amino acid substrate
H
OP
O
C
H
N
OP
C
O
N
OH
N
CH3
enz
+
NH 3
OH
transimination
N
H
CH3
H
enzyme-PLP adduct
(external aldimine)
substrate-PLP adduct
(internal aldimine)
fig 3
With the preliminary transimination accomplished, the real PLP chemistry is ready to
start. The versatility of PLP in terms of its ability to assist with a wide variety of reaction
types is illustrated by the figure below, showing how, depending upon the
reaction/enzyme in question, PLP can assist in the cleavage of any one of the four bonds
to the α-carbon of the amino acid substrate.
decarboxylation
racemization, elimination,
substitution
H
O
R C C O
N
retro-aldol, retro-Claisen
transamination
OH
PO
N
CH3
H
fig 4
Let's look first at the reaction catalyzed by PLP-dependent alanine racemase. (EC
5.1.1.1).
17.1B: PLP-dependent amino acid racemization
In section 12.2B we saw an example of a PLP-independent amino acid racemization
reaction, in which the negatively-charged intermediate was simply the enolate form of a
carboxylate:
354
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
R
H R
O
H3N
H R
O
H3N
O
O
O
enolate intermediate
L-amino acid
O
H3N
D-amino acid
fig 5
Many other amino acid racemase reactions, however, require the participation of PLP.
Like all other PLP-dependent reactions that we will see in this section, PLP-dependent
amino acid racemization begins with a preliminary step in which the substrate becomes
attached to the coenzyme through a transimination. Once it is linked to PLP in the active
site, the α-proton of an amino acid substrate is abstracted by an active site base (step 1
below). The negative charge on the carbanion intermediate can, of course, be delocalized
to the carboxylate group. The PLP coenzyme, however, provides an expanded network
of conjugated π-bonds over which the electron density can be delocalized all the way
down to the PLP nitrogen. This is what we mean when we say that the job of PLP is to
act as an ‘electron sink’: the coenzyme is very efficient at absorbing, or delocalizing, the
excess electron density on the deprotonated α-carbon of the reaction intermediate. PLP
is helping the enzyme to increase the acidity of the α-hydrogen by stabilizing the
conjugate base. A PLP-stabilized carbanion intermediate is commonly referred to as a
quinonoid intermediate. Note that in the overall reaction equation below, PLP appears
below the reaction arrow in brackets, indicating that it participates in the mechanism but
is regenerated as part of the reaction cycle.
PLP-dependent amino acid racemization:
H R
H R
O
H3N
[PLP]
O
L-amino acid
O
H3N
O
D-amino acid
term in brackets means that the coenzme participates in
catalysis but is regenerated as part of the reaction cycle
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
355
Chapter 17: Coenzymes
Mechanism:
Preliminary step - transimination:
O
R
O
NH 3
enz
R
amino acid substrate
H
OP
O
H
N
C
OP
C
O
N
OH
N
enz
+
NH 3
OH
transimination
CH3
N
H
CH3
H
enzyme-PLP adduct
(external aldimine)
substrate-PLP adduct
(internal aldimine)
First step - deprotonation:
base positioned behind substrate
B:
R
H
R
CO 2
N
OP
CO 2
N
OP
OH
R
CO 2
N
OP
OH
OH
step 1
N
CH3
H
substrate-PLP adduct
356
N
CH3
H
N
H
quinonoid intermediate
(PLP-stabilized carbanion)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
CH3
Chapter 17: Coenzymes
Second step - reprotonation from the opposite side:
acid positioned in front of substrate
A
H
R
R
CO 2
H
N
OP
N
OP
OH
N
CH3
CO 2
OH
step 2
N
H
CH3
H
product-PLP adduct
Final step - transimination:
enz
R
H
CO2
H
NH3
N
OP
R
enz
CO2
NH3
N
OP
OH
OH
imine exchange
N
CH3
N
CH3
H
H
product-PLP adduct
enzyme-PLP adduct
fig 7
fig 6 fig 6a
Just as in the PLP-independent racemase reactions, reprotonation occurs on the opposite
side of the substrate (step 2), leading to the D-amino acid product.
All that remains is the final imine exchange which frees the D-amino acid product and reattaches the coenzyme to the enzymatic lysine side-chain, ready to begin another catalytic
cycle.
To simplify matters, from here on we will not include the preliminary and final
transimination steps in our PLP reaction figures - we will only show mechanistic steps
that occur while the substrate is attached to the coenzyme (the internal aldimine forms).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
357
Chapter 17: Coenzymes
17.1C: PLP-dependent decarboxylation
In the amino acid racemase reaction above, PLP assisted in breaking the α-carbon to αproton bond of the amino acid. Other PLP-dependent enzymes can catalyze the breaking
of the bond between the α-carbon and the carboxylate carbon by stabilizing the resulting
carbanion intermediate: these are simply decarboxylation reactions.
PLP-depended amino acid decarboxylation:
O
R
CO2
R
O
NH3
[PLP]
NH3
Mechanism:
H A
O
R
N
OP
R
R
O
H
N
OP
OH
N
OP
OH
OH
step 1
N
CH3
H
substrate-PLP adduct
N
N
CH3
H
H
quinonoid intermediate
step 2
R
N
OP
OH
N
CH3
H
product-PLP adduct
fig 8 fig 9
358
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
CH3
+ CO2
Chapter 17: Coenzymes
Notice something very important here: while in racemization reactions the assistance of
PLP can be seen as 'optional' (in the sense that some racemase enzyme use PLP and
others do not), the coenzyme is essential for amino acid decarboxylation steps. Without
PLP, there is no way to stabilize the carbanion intermediate, and decarboxylation is not a
chemically reasonable step.
H
R C
negative charge on carbon
cannot be stabilized
O
NH3
R
O
X
C
H
NH3
fig 10
One example of a PLP-facilitated decarboxylation reaction is the final step in the lysine
biosynthesis pathway: (EC 4.1.1.20).
NH3
NH3
CO2
CO2
CO2
O
H3N
[PLP]
H3N
lysine
O
fig 11
Exercise 17.1: Draw mechanistic arrows for the carbon-carbon bond-breaking step of the
PLP-dependent decarboxylation reaction above.
17.1D: PLP-dependent retroaldol and retro-Claisen cleavage
(It would be a good idea before reading this section to review aldol/retro-aldol and
Claisen/retro-Claisen reaction mechanisms in sections 12.3 and 13.3, respectively)
So far we have seen PLP playing a role in breaking the bond between the α-carbon and
its α-proton (in the racemization reaction), and the bond between the α-carbon and
carboxylate carbon (in the decarboxylation reaction).
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
359
Chapter 17: Coenzymes
decarboxylation
racemization, elimination,
substitution
H
O
R C C O
N
retro-aldol, retro-Claisen
transamination
OH
PO
N
CH3
H
fig 4
Other PLP-dependant enzymes catalyze cleavage of the bond between the α-carbon and
the first carbon on the amino acid side chain, otherwise known as the β-carbon. In the
serine degradation pathway, serine is first converted to glycine by a retro-aldol cleavage
reaction. (). Although a reasonable mechanism could be proposed without the
participation of PLP, this reaction in fact requires the coenzyme to assist in stabilization
of the negative charge on the carbanion intermediate.
360
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
A PLP-dependent retro-aldol cleavage reaction
(serine hydroxymethyl transferase, EC 2.1.2.1)
O
H2C
H
H
H
CO2
+
[PLP]
NH3
O
CO2
H
H
NH3
C
H
β-carbon
glycine
serine
formaldehyde
Mechanism:
:B
H
O
O
H
C
H
CO 2
H
H
H
CO 2
N
OP
OH
OH
step 2
step 1
CH3
C
N
OP
OH
N
H
CO 2
N
OP
A
N
CH3
H
N
CH3
H
glycine-PLP adduct
serine-PLP adduct
fig 12 fig 13
Note that, in this reaction just as in the racemase reaction described previously, the key
intermediate is a PLP-stabilized carbanion, or quinonoid.
What happens to the (toxic!) formaldehyde produced in this reaction? We will see later
in this chapter how the serine hydroxymethyltransferase enzyme goes on to use another
coenzyme called tetrahydrofolate to prevent the formaldehyde from leaving the active
site and causing damage to the cell.
PLP also assists in retro-Claisen cleavage reactions (section 13.3C), such as this step in
the degradation of threonine. (EC 2.3.1.29)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
361
Chapter 17: Coenzymes
A PLP-dependent retro-Claisen reaction:
CH3
CoASH
CO2
CO2
O
NH3
[PLP]
CH3
+
NH3
O
SCoA
glycine
acetyl CoA
Mechanism:
:B
CH3
H
SCoA
CH3
CO2
O
O
N
OP
SCoA
CO2
H3C
N
OP
OH
CH3
H
SCoA
H A
CO2
N
OPH2
OH
OH
step 1b
step 1a
N
O
N
CH3
N
H
substrate-PLP adduct
CH3
H
quinonoid intermediate
step 2
CO2
N
OP
OH
N
CH3
H
glycine-PLP adduct
fig 14 fig 15
Notice how, like the retro-aldol reaction, the bond between the α-carbon and the βcarbon of the amino acid substrate is broken (in step 1b).
362
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
17.1-E: PLP-dependent transamination
One of the most important reaction types in amino acid metabolism is transamination, in
which an amino group on a donor molecule (often an amino acid) is transferred to a
ketone or aldehyde acceptor molecule.
A transamination reaction:
R1
CO2
R3
R2
+
NH3
O
amino acid substrate
ketone or aldehyde
CO2
R1
[PLP]
+
R3
R2
O
NH3
α-keto acid
amine
Transamination phase 1 (transfer of amino group from amino acid substrate to
coenzyme)
R1
CO2
amino group has been
transferred to PLP
N
OP
H2O
NH3
PO
OH
OH
N
N
CH3
+
CH3
H
H
substrate-PLP adduct
CO2
R1
O
α-keto acid
pyridoxamine phosphate
(PMP)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
363
Chapter 17: Coenzymes
Mechanism:
position of imine has shifted
B:
H
R
OP
H
CO2
R
N
OP
H
N
OH
H
CH3
CO2
H
H A
PO
N
H
OH
OH
step 2
step 1
N
R
CO2
N
CH3
N
H
CH3
H
H2O
amino group has been
transferred to PLP
H
PO
step 3
(imine hydrolysis)
NH3
H
OH
N
+
CH3
H
pyridoxamine phosphate
(PMP)
CO2
R
O
α-keto acid
fig 16 fig 17 fig17a
Once again, step 1 is abstraction of the α-proton from the PLP-substrate adduct.
However, in a transaminase reaction this initial deprotonation step is immediately
followed by reprotonation at what was originally the aldehyde carbon of PLP (step 2
above), which results in a new carbon-nitrogen double bond (in other words, an imine)
between the α-carbon and the nitrogen atom of the original amino acid. The repositioned
imine group is then hydrolyzed (step 3 above), breaking the carbon-nitrogen bond,
transferring the amino group to the coenzyme, and releasing an α-keto acid.
The coenzyme, which now carries an amine group and is called pyridoxamine phospate
(PMP), next transfers the amine group to α-ketoglutarate (to form glutamate) through a
reversal of the whole process depicted above.
364
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
Transamination reaction, phase 2
(transfer of amino group from coenzyme to acceptor molecule)
R2
NH3
PO
OH
N
R3
R2
+
H2O
R3
N
OP
OH
O
CH3
N
H
CH3
H
pyridoxamine phosphate
(PMP)
Mechanism:
see exercise below
fig 18
In a transamination reaction, the PLP coenzyme not only provides an electron sink, it also
serves as a temporary 'parking place' for an amino group as it is transferred from donor to
acceptor.
Exercise 17.2: Show a complete, step-by-step mechanism for 'phase 2' of the
transamination reaction above.
Here is an example of a transamination reaction in the arginine biosynthesis pathway: EC
2.6.1.11
O
CO2
CO2
H3C
O
NH
+
CO2
NH3
H3C
CO2
CO2
[PLP]
NH
+
CO2
O
O
NH3
fig 19
Exercise 17.3:
a) Draw arrows for the first mechanistic step of 'phase 2' of the above transaminase
reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
365
Chapter 17: Coenzymes
b) Which carbon on the substrate side of the reaction will eventually become the αcarbon of arginine?
Exercise 17.4: Propose a pathway, with three enzymatic steps, for the biosynthesis of
serine from 3-phosphoglycerate. Include a generalized enzyme name (for example,
kinase, methyltransferase, aldolase, etc.) for each step. Glutamate plays a role in the
process as an amino group donor.
OH
OP
HO
CO2
3-phosphoglycerate
H3N
CO2
serine
fig 20
17.1F: PLP-dependent β-elimination and β-substitution
(Before starting this section, it would be a good idea to review E1cb β-elimination and
conjugate addition reaction mechanisms in chapter 13.4)
By now it should be pretty apparent that PLP is a pretty versatile coenzyme! Two more
reaction types in the PLP toolbox are β-elimination and β-substitution on amino acid
substrates.
366
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
In a PLP-dependent β-elimination reaction, the coenzyme simply helps to stabilize the
carbanion intermediate of the E1cb mechanism:
A PLP-dependent β-elimination reaction
X O
XH
O
β
H
O
α
O
[PLP]
NH3
NH3
Mechanism:
X
B:
X
N
OP
N
OP
N
OP
OH
OH
OH
N
CO2
CO2
CO2
H
CH3
step 2
step 1
N
N
CH3
H
H
H
CH3
substrate-PLP adduct
product-PLP adduct
fig 21 fig 22
Serine dehydratase (EC 4.2.1.13) catalyzes a PLP-dependent β-elimination in the first
step of the serine degradation pathway:
OH
H2O
O
H
O
NH3
O
O
H3C
O
[PLP]
NH3
tautomerization
O
NH2
serine
H2O
PLP-dependent phase of the reaction
imine hydrolysis
NH4+
O
H3C
O
O
pyruvate
fig 23
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
367
Chapter 17: Coenzymes
A β-substitution reaction is simply E1cb elimination followed directly by the reverse
reaction (conjugate addition) with a different nucleophile (Y in the figure below):
A β-substitution reaction:
X
Y
CO2
H
CO2
N
OP
XH
YH
N
OP
OH
OH
N
CH3
H
N
OP
OH
conjugate addition
β-elimination
N
CO2
H
CH3
N
H
CH3
H
fig 24
In many bacteria, the synthesis of cysteine from serine includes a PLP-dependent βsubstitution step (EC 2.5.1.47).
O
O
H3C
O
H
HS-
H3C
CO2
NH3
acetylserine
SH
OH
CO2
H
[PLP]
NH3
cysteine
fig 25
Exercise 17.5: Draw a mechanism for the conjugate addition phase of the reaction above
(end with the cysteine-PLP adduct).
17.1G: PLP-dependent γ-elimination and γ-substitution reactions
The electron sink capability of PLP allows some enzymes to catalyze eliminations at the
γ-carbon of some amino acid side chains, rather than at the β-carbon. The secret to
understanding the mechanism of a γ-elimination is that PLP essentially acts as an electron
sink twice - it absorbs the excess electron density from not one but two proton
abstractions.
368
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
PLP-dependent γ-elimination
X
γ
XH
CO2
β
CO2
[PLP]
α
NH3
NH3
Mechanism:
2nd proton abstraction from β-carbon
X
γ
B:
B:
β
H
CO2
β
H
α
N
OP
X
γ
α
N
OP
CH3
CO2
CO2
H
N H
OP
O
OH
N
X
step 1
N
CH3
O
step 2
N
H
H
CH3
H
step 3
A
H
CO2
CO2
N
OP
N
OP
OH
OH
step 4
N
CH3
H
N
CH3
H
fig 26 fig 27
In a familiar first step, the α-proton of the amino acid is abstracted by an enzymatic base,
and the electron density is absorbed by PLP. Next comes the new part - before anything
happens to the electron density from the first proton abstraction, a second proton, this
time from the β-carbon on the side chain, is abstracted, forming an enamine intermediate
(step 2). The phenolic proton on the pyridoxal ring of PLP donates a proton to the
nitrogen. In step 3, the leaving is expelled and a new π-bond forms between the β and γ
carbons (step 3). This π-bond is short-lived, however, as the electron density from the
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
369
Chapter 17: Coenzymes
first proton abstraction, which has been 'stored' in PLP all this time, flows back up to
protonate the α-carbon (step 4), leaving the γ-elimination product linked to PLP via the
usual imine connection.
An example is the cystathionine γ-lyase reaction in the methionine degradation pathway
(EC 4.4.1.1):
NH3
CO2
S
NH3
CO2
NH3
[PLP]
CO2
NH3
+
CO2
SH
cysteine
fig 28
370
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
A related reaction is PLP-dependent γ-substitution, which again is simply γ-elimination
of a leaving group (X in the figure below) followed directly by the reverse process (a γaddition) with a different nucleophile ('Nu' in the figure below).
PLP-dependent γ-substitution:
X
γ
YH
XH
CO2
β
α
Y
CO2
[PLP]
NH3
NH3
Mechanism:
:B
H
Y
Y
Y
A
CO2
N
OP
N
CO2
step 4
O
N
CH3
H
A
N H
OP
H
O
H
CO2
H
N
OP
step 5
OH
N
CH3
CH3
H
H
step 6
steps 1, 2, 3 from fig 27
X
Y
CO2
CO2
N
OP
N
OP
OH
N
OH
CH3
N
H
CH3
H
fig 29 fig 30
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
371
Chapter 17: Coenzymes
Below is a PLP-dependent γ-substitution reaction in the methionine degradation pathway
(EC 4.2.1.22):
NH3
O
CO2
CO2
O
S
NH3
CH2O2
NH3
+
CO2
SH
cysteine
[PLP]
CO2
+
O
CO2
O
NH3
fig 31
17.1H: Racemase to aldolase: altering the course of a PLP reaction
We have seen how PLP-dependent enzymes catalyze a variety of reaction types racemization, retroaldol/retro-Claisen cleavage, transaminination, elimination, and
substitution - which, despite their apparent diversity, are all characterized by formation of
a critical carbanion intermediate which is stabilized by the 'electron sink' property of the
PLP coenzyme. Given this common mechanistic feature, it would be reasonable to
propose that the active site architecture of these enzymes might also be quite close. This
idea was nicely illustrated by an experiment in which researchers found that changing a
single active site amino acid of PLP-dependent alanine racemase was sufficient to turn it
into a retro-aldolase (J. Am. Chem. Soc. 2003, 125, 10158).
In the 'wild-type' (natural) alanine racemase reaction, an active site histidine (red in
figure below) deprotonates a neighboring tyrosine residue (blue), which in turn acts as the
catalytic base abstracting the α-proton of the substrate. When researchers changed the
tyrosine to an alanine (using a technique called 'site-directed mutagenesis'), and
substituted β-hydroxytyrosine for the alanine substrate, the new 'mutant' enzyme
catalyzed a retro-aldol reaction.
372
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
enz
enz
racemase
(wild-type)
O
HN
N
H3C
H
H
CO2
N
OP
OH
N
CH3
H
retroaldolase
(mutant)
H3C
enz
HN
N
H
enz
O
CO2
H
N
OP
OH
N
CH3
H
fig 32
Notice what has happened here: the basic histidine, with no tyrosine to deprotonate
because of the mutation, is instead positioned to abstract a proton from the β-hydroxyl
group of the new substrate, setting up a retroaldol cleavage. That was all it took to change
a racemase into a retroaldolase, because the necessary PLP electron sink system was all
left in place. The researchers predicted correctly that the phenyl ring of β-hydroxy
tyrosine would fit nicely in the space left empty due to the tyrosine to alanine change in
the mutant enzyme's structure
These results underline the close mechanistic relationship between two PLP-dependent
reactions which, at first glance, appear to be quite different - and suggest that PLPdependent racemases and aldolases may have evolved from a common 'ancestor' enzyme.
17.1I: Stereoelectronic considerations of PLP-dependent reactions
Recall that all PLP-dependent reactions involve the cleavage of one of the bonds coming
from the α-carbon of an amino acid substrate, with the coenzyme serving as an 'electron
sink' to stabilize the intermediate that results. PLP-dependent enzymes accelerate this
bond-breaking step by binding the substrate-PLP adduct in a conformation such that the
bond being broken is close to perpendicular to the plane formed by the conjugated π
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
373
Chapter 17: Coenzymes
system of PLP: this way, as the α-carbon transitions from sp3 to sp2 hybridization, the
unhybridized p orbital is already oriented to overlap with the rest of the conjugated
system. For example, in alanine racemase the first step is cleavage of the Cα-H bond, so
it must be that bond which is positioned near-perpendicular to the PLP plane:
racemase
breaking bond is oriented close to
perpendicular with delocalized π system of PLP
B:
HO
H
H
CH3
N
H3C
N
H3C
PO
PO
CH3
HO
H
N
O2C
N
O2C
H3C
N
H3C
CH3
HO
H
H
CH3
NH
O2C
NH
O2C
HO
H
PO
N
PO
Likewise, in an amino acid decarboxylase, the Cα-carboxylate bond is held nearperpendicular to the PLP plane, and in hydroxymethyltransferase, the Cα-Cβ bond is in
the perpendicular orientation:
serine hydroxymethyltransferase
decarboxylase
B:
O
O
H3C
H
H
HO
PO
O
CH3
NH
N
H
H
HO
CH3
NH
H
O2C
N
PO
These are all good examples of how enzymatic catalysis is achieved, in part, by the
ability of the active site to bind the substrate molecule in a specific conformation which
contributes to the lowering of the activation energy of a key reaction step.
Section 17.2: Thiamine diphosphate (Vitamin B1)
Thiamine diphosphate (ThDP, sometimes also abbreviated TPP or ThPP) is a coenzyme
which, like PLP, acts as an electron sink to stabilize key carbanion intermediates. The
most important part of the ThDP molecule from a catalytic standpoint is its thiazole ring.
374
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
thiazole ring
H
NH2
N
N
OPP
N
H3C
S
H3C
thiamine diphosphate (ThDP)
fig 33
The proton on the carbon between nitrogen and sulfur on the thiazole ring is weakly
acidic, with a pKa of about 18.
H
N
S
R
N
S
H3C
R
H3C
R
R
ThDP ylide
fig 34
The reason for its acidity lies partly in the ability of the neighboring sulfur atom to
accept, in its open d-orbitals, some of the excess electron density of the conjugate base.
Another reason is that the positive charge on the nitrogen helps to stabilize the negative
charge on the conjugate base. The deprotonated thiazole is called an ylide, which is a
general term for a species with adjacent positively and negatively charged atoms.
The negatively charged carbon on the ylide form of ThDP is nucleophilic, and as we shall
soon see, the first step of most TPP-dependent reactions is nucleophilic attack of the ylide
carbon on a carbonyl group of the reaction substrate.
ThDP plays a key role in a variety of reaction types, but the common theme in all ThDPdependent reactions is cleavage of a bond adjacent to the carbonyl carbon of a ketone or
aldehyde.
Thiamine diphosphate assists in breaking bonds next to a ketone or aldehyde:
O
R
O
O
H
R
R
O
R
OH
O
fig 35
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
375
Chapter 17: Coenzymes
Consider this hypothetical decarboxylation step:
O
R
C
C
O
O
X
R
O
localized charge on carbonyl
C carbon - highly unstable
fig 37
Hopefully, you can quickly recognize that this is not a chemically reasonable step,
because the intermediate species which results from decarboxylation has a negative
charge localized on the ketone carbon - a very unstable, unlikely intermediate indeed.
(Recall from section 13.1 that decarboxylation steps usually result in intermediates in
which the negative formal charge is delocalized to an oxygen or nitrogen - in other
words, enolates or enamines.)
Now consider, however, a reaction going on in your cells right now, catalyzed by the
enzyme pyruvate decarboxylase (EC 4.1.1.1):
H+
CO2
O
O O
H3C C C O
pyruvate
[ThDP]
H3C C
H
acetaldehyde
fig 36
Somehow, the enzyme manages to accomplish this 'impossible' decarboxylation. How
does this happen? Here is where the thiamine diphosphate coenzyme comes in.
376
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
A ThDP-dependent decarboxylation reaction (pyruvate decarboxylase):
CO2
H+
O
O O
R C C O
R C
[ThDP]
H
Mechanism:
H A
R N
S
H3C
R
O
O O
O O
R C C O
H
H
R C C O
step 1
H A
R C
step 2
R N
S
R N
S
H3C
R
H3C
R
step 3
B:
H
O
O
R C
R N
S
H3C
R
H
step 4
R C
H
R N
S
H3C
R
fig 36a fig 38
Upon binding to the enzyme's active site, ThDP quickly loses a proton. The nucleophilic
ylide carbon then adds to the carbonyl carbon of pyruvate.
Look carefully at the intermediate that results from step 1 in the mechanism above. The
thiazole ring of ThDP, once it has added to the carbonyl of pyruvate, provides an
'electron sink' to absorb the electrons from decarboxylation (step 2). Note which bond is
breaking in step 2 - as was mentioned earlier, the common function of ThDP is to make
possible the cleavage of a bond to a carbonyl carbon.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
377
Chapter 17: Coenzymes
In step 3, the electrons from decarboxylation flow back to abstract a proton from an
acidic group in the active site. All that remains is for the product to break free of
thiamine in step 4.
Thiamine can also assist in decarboxylation-addition reactions:
ThDP-dependent decarboxylation-addition:
O O
R C C O
+
C
H
CO2
H+
O
O OH
R C C R
R
ThDP]
H
Mechanism:
A
H
O
H A
H
R C C O
R N
S
H3C
R
R C C O
step 1
C
R
O
O O
O O
H
H
R C
step 2
R N
S
R N
S
H3C
R
H3C
R
step 3
B:
H
O OH
R C C R
O
step 4
H
R N
S
H3C
R
R C
OH
C R
R N
H
S
H3C
R
fig 39a fig 39
Here, the electron-rich intermediate formed from the decarboxylation step (step 2) simply
goes on to act as a nucleophile rather than as a base, adding to the carbonyl group of an
aldehyde or ketone (step 3). As before, the product breaks free of ThDP in step 4.
378
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
An example is the first step in the biosynthetic pathway for isoprenoid compounds in
bacteria:
O
O
H3C
O
+
H
CO2
H+
OH
OP
O
[ThDP]
O
OH
OP
H3C
OH
fig 40
Transketolase, a ThDP-dependent enzyme in the pentose phosphate pathway of sugar
metabolism, catalyzes a carbon-carbon bond break step, followed by a carbon-carbon
bond forming step. The substrates and products are at similar energy levels, so the
reaction is completely reversible.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
379
Chapter 17: Coenzymes
Transketolase reaction:
R C C R1
H
+
H
C
O
O OH
O
O OH
+
R C C R2
[ThDP]
R2
H
H
C
R1
Mechanism:
A
O
B:
H A
O O
H
H
H
R C C R1
H
R N
S
H3C
R
O O
step 1
H
R C C R1
H
H
R C
step 2
O
H
S
R N
S
H3C
R
H3C
R
step 3
B:
H
O OH
H
R N
H3C
O
R C
step 4
OH
C R
R N
H
S
H3C
R
S
R
fig 41 fig 42
380
R2
O
R N
R C C R
C
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
H
C
R1
Chapter 17: Coenzymes
Below is an actual example of a transketolase-catalyzed transformation from the pentose
phosphate pathway (shown in Fischer projections, as is common for sugar structures).
OH
O
OH
HO
O
O
OH
HO
+
OH
OH
OH
OH
OP
OP
O
[ThDP]
ribose-5phosphate
xylulose-5phosphate
OH
+
OH
OH
OP
OP
glyceraldehyde-3phosphate
sedoheptulose-7phosphate
fig 43
Exercise 17.6: As was mentioned above, the transketolase reaction is highly reversible.
Do you think the same can be said for the decarboxylation and decarboxylation-addition
reactions we saw in this section? Why or why not?
Exercise 17.7: Propose a mechanism for the reaction below. J. Mol. Biol. 2010, 401, 253
CO2
OH
O
O2C
CO2
pyruvate
CO2
+
O
CO2
OH
O2C
O
CO2
[ThDP]
fig 44
Exercise 17.8: Propose a mechanism for the reaction below. (Hint - in this case, ThDP
allows for an E1cb-type β-elimination step that otherwise would be impossible).
O
O
R
[ThDP]
OH
O
R
fig 44a
Section 17.3: Thiamine diphosphate, lipoamide and the pyruvate dehydrogenase
reaction
The enzyme pyruvate dehydrogenase is one of the most central of all the enzymes of
central metabolism: by converting pyruvate to acetyl-CoA, it links glycolysis (where
glucose is broken down into pyruvate) to the citric acid cycle, into which carbons enter in
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
381
Chapter 17: Coenzymes
the form of acetyl-CoA. Five ceonzymes are involved: coenzyme A, nicotinamide,
thiamine diphosphate, FAD, and finally lipoamide, one which is new to us at this point.
Reaction catalyzed by pyruvate dehydrogenase:
HSCoA
NAD+
O
CO2
NADH
O
H3C
O
pyruvate
[ThDP]
[lipoamide]
[FAD]
O
H3C
SCoA
acetyl CoA
fig 45
You will learn more about the structure and metabolic role of this complex and
remarkable enzyme in a biochemistry course. Here, we will focus on the multi-step
organic reaction it catalyzes, which we are at long last equipped to understand.
Looking at the reaction, you should recognize that, first of all, the pyruvate substrate is
being oxidized - it starts out as a ketone, and ends up as a thioester, losing carbon dioxide
in the process. Ultimately, the oxidizing agent in this reaction is NAD+, but the reduction
of NAD+ is linked to the oxidative decarboxylation of pyruvate by FAD and a disulfidecontaining coenzyme called lipoamide, which is lipoic acid attached by an amide linkage
to a lysine residue on the enzyme.
S
OH
S
O
lipoic acid
lysine residue in active site
reactive part of coenzyme
O
H
N
S
enz
S
O
HN
enz
lipoamide
(linked to lysine residue in enzyme)
fig 45a
382
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
The second thing to notice is that, because the reaction involves breaking the bond
between the ketone carbon and an adjacent carbon, thiamine diphosphate (ThDP)
coenzyme is required. In fact, the first phase of the reaction (steps 1 and 2 below) is
identical to that of pyruvate decarboxylase, an enzyme we discussed a few pages ago.
The pyruvate decarboxylase reaction mechanism:
Phase 1: Decarboxylation of pyruvate
H A
O
O
H3C
O
R N
S
H3C
R
H O
step 1
H
O
H3C
O
H3C
O
step 2
R N
S
R N
S
H3C
R
H3C
R
fig 46
The ThDP-stabilized carbanion then acts as a nucleophile, cleaving the disulfide bridge
of lipoamide (step 3 below). It is in this step that oxidation of the substrate is actually
occurring. After the resulting thioester product is released from ThDP (step 4 below), it
undergoes transesterification form acetyl-CoA, the product of the reaction.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
383
Chapter 17: Coenzymes
Phase 2 of the pyruvate decarboxylase reaction mechanism: lipoamide-mediate
oxidation to acetyl-CoA
A
R
H
S
H
H 3C
B:
S
O
O
R N
S
H 3C
R
H
H 3C
step 3
thioester
S
SH
O
R
R N
S
H 3C
R
S
SH
CH3
step 4
R
HSCoA
transthioesterification
(see section 11.5A for mechanism)
O
HS
SH
R
+
H 3C
SCoA
acetyl-CoA
dihydrolipoamide
(reduced)
fig 47
We are not done yet! In order for the catalytic cycle to be complete, the reduced
dihydrolipoamide must be regenerated back to its oxidized state through disulfide
exchange with a disulfide bond on the enzyme. The pair of enzymatic cysteines is then
oxidized back to disulfide form by an FAD-dependent reaction.
384
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
Phase 3 of the pyruvate decarboxylase reaction mechanism:: regeneration of
lipoamide
R
R
HS
S
HS
S
dihydrolipoamide
lipoamide
+
+
S
disulfide exchange
(see section 15.6 for mechanism)
S
enz
SH HS
enz
enz
enz
FAD
oxidation of free cysteines by FADH
(see section 15.6 for mechanism)
FADH 2
S
S
enz
enz
fig 48
Finally, FAD is regenerated with concurrent reduction of NAD+:
Phase 4: Regeneration of FADH2:
H
O
N
N
N
N
R
H
H
O
NAD +
NADH
(see section 15.4 for mechanism)
O
N
N
N
N
H
O
R
FAD
FADH2
fig 49
Section 17.4: Folate
Folate, or vitamin B9, is essential for a variety of important reactions in nucleotide and
amino acid metabolism. The reactive part of folate is the pterin ring system, shown in
red below. The conventional atom numbering system for folate is also indicated.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
385
Chapter 17: Coenzymes
H2N
1
8
N 1a N
2
6
N
H
7
3 4
4a
O
9
N
5
10
N
H
H
N
pterin ring system
O
folate
CO2
CO2
fig 50
17.4A: Active forms of folate
Folate is active as a coenzyme in its reduced forms, dihydrofolate and tetrahydrofolate,
which are formed by NADPH-dependent imine hydrogenation steps (section 15.3).
H
H2N
H
N
N
N
H
NADPH, H+
folate
H2N
H
N
O
NADP+
H
N
N
N
H
N
O
R
H
N
H
N
R
dihydrofolate (DHF)
NADPH, H+
NADP+
H
H2N
H
N
N
H
N
H
H
N
O
H
H
N
R
tetrahydrofolate (THF)
fig 51
The metabolic role of folate is to serve as a donor or acceptor in single-carbon transfer
reactions. How is folate different from S-adenosylmethionine (SAM, section 8.8A)
which also serves as a single-carbon donor? Recall that SAM donates a single carbon in
the form of a methyl group: essentially, the single carbon of SAM is in the methanol
386
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
(CH3OH) oxidation state, because it has only one bond to a heteroatom (specifically, to
sulfur). (Refer to section 15.1 for a review of oxidation states).
NH2
NH3
O2C
N
S
H3C
carbon is in the methanol oxidation state
(one bond to an electronegative heteroatom)
O
N
N
N
OH OH
S-adenosylmethionine
(SAM)
fig 52a
Folate coenzymes, on the other hand, can carry a single carbon in the formaldehyde and
formate oxidation states, in addition to the methanol oxidation state. By 'formaldehyde'
and 'formate' oxidation state, we mean that the carbon has two and three bonds to
heteroatoms, respectively.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
387
Chapter 17: Coenzymes
H
H2N
N
N
H
O
N
N
H
H C
N
carbon is in the formate oxidation state
(three bonds to electronegative heteroatoms)
R
O
H C
O
O
formyltetrahydrofolate
(f-THF)
formate
H
H2N
H
N
N
N
carbon is in the formaldehyde oxidation state
(two bonds to electronegative heteroatoms)
N
O H C
H
N
R
H C
methylenetetrahydrofolate
(CH2-THF)
H
O
formaldehyde
H
H2N
H
N
N
N
N
O
CH3 HN
methyltetrahydrofolate
(CH3-THF)
carbon is in the methanol oxidation state
(one bond to an electronegative heteroatom)
R
H
H C OH
H
methanol
fig 52
Some key reactions in nucleic acid and amino acid metabolic pathways involve transfer
of a single carbon in the formaldehyde or formate states. However, this could present
problems. Formaldehyde by itself is very toxic: in particular, it tends to spontaneously
form unwanted crosslinks between amine groups (eg. lysine side chains) in proteins.
388
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
proteins are now crosslinked by formaldehyde
O
H
C
H2O
OH
H C
H
H C
H
NH
protein
protein
NH2
NH2
NH2
protein
protein
protein
protein
H
NH
NH
fig 53
Free formaldehyde is too reactive, and would cause damage to a cell. The CH2-THF
coenzyme is stable in solution, but in the active site of certain enzymes it is reactive
enough to serve as a formaldehyde donor, as we will see shortly.
Free formate, on the other hand, is a carboxylate, and we know from chapter 11 that
carboxylates are not reactive in acyl substitution steps. Formate could be activated by
phosphorylation, of course, but the resulting formyl phosphate would be too reactive in
many enzyme active sites. A 'happy medium' has been found in which carbons in the
formate oxidation state are carried by folate in the form of f-THF: once again, the carbon
donor is stable in solution, but sufficiently reactive in certain enzyme active sites to
accomplish controlled transfer of a formate group.
17.4B: Formation of formyl-THF and methylene-THF
Formyltetrahydrofolate (f-THF) is formed from THF and a formate molecule which has
been activated by phosphorylation (formyl phosphate, as stated in the paragraph above, is
a high reactive intermediate, but is held inside the enzyme's active site for immediate
reaction with the incoming amine group of THF).
O
H
C
O
ATP
H
H2N
H
N
N
O
Pi
O
+
N
H
ADP
N
H
N
H
H
C
OP
H2N
H
R
THF
N
N
O
N
N
H
H C
f-THF
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
N
R
O
389
Chapter 17: Coenzymes
fig 54
There are two main metabolic routes to CH2-THF. One route is just the last step of the
serine hydroxymethyltransferase reaction we have already seen in section 17.1D: the
formaldehyde formed in the PLP-dependent phase of the reaction stays in the active site,
and the oxygen is displaced by successive attacks from the amine nucleophiles at the 5
and 10 positions of THF. (Notice the similarity to the formaldehyde-protein crosslinking
reaction shown earlier in this section.)
OH
H
C
H
H
C
H 3N
CO 2
serine
H
H 3N
H 2N
H
N
H
N
O
H
N
O
CO 2
pyruvate
+
N
H
C
H
N
H
C
H 2O
H 2N
H
formaldehyde
R
THF
H
N
N
H
N
N
O H C N
R
H
CH2-THF
fig 55
In the second route, f-THF is dehydrated, then the resulting methenyl-THF intermediate
is reduced by NADPH.
390
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
H2N
H
H
N
N
N
H2O
H
N
H
O
H2N
H C
N
H
N
N
N
N
C
O
R
N
R
H
O
methenyltetrahydrofolate
(CH-THF)
formyltetrahydrofolate
(f-THF)
NADPH
NADP+
H2N
H
H
N
N
N
N
O H C
H
N
R
methylenetetrahydrofolate
(CH2-THF)
fig 56
Methylene-THF (CH2-THF) is reduced to methyl-THF (CH3-THF) in a flavin-dependent
reaction. Biochemistry 2001, 40, 6216
H2N
H
N
N
H
N
NADPH
N
O H C
H
NADP+
[FADH2]
N
H2N
H
N
N
methylenetetrahydrofolate
(CH2-THF)
N
O
R
H
N
CH3 HN
R
methyltetrahydrofolate
(CH3-THF)
fig 56a
Exercise 17.9: Draw a likely mechanism for the reduction of CH2-THF to CH3-THF by
FADH2.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
391
Chapter 17: Coenzymes
Recall from the introduction to this chapter that babies who were in the womb during the
Dutch Hunger Winter faced a higher risk, when they reached adulthood, of conditions
such as obesity and schizophrenia, most likely caused by disruptions in the Sadenosylmethionine-dependent methylation of their DNA, which was in turn caused by
their mothers not getting enough folate. CH3-THF is the source of the methyl group in
methionine, which ultimately becomes the methyl group in SAM. The methylation of
homocysteine to methionine (below) involves a cobalt-containing coenzyme called
cobalamin, but the mechanism for this reaction is beyond the scope of our discussion.
The second reaction below (formation of SAM) is simply an SN2 displacement of the
inorganic triphosphate leaving group on ATP by the nucleophilic sulfur in methionine.
(metallo-enzyme reaction mechanism is beyond our scope)
NH3
CH3-THF
NH3
THF
O2C
O2C
SH
[cobalamin]
S
H3C
homocysteine
methionine
ATP
(SN2 displacement of
inorganic triphosphate)
PPPi
NH2
NH3
O2C
N
S
H3C
O
N
N
N
OH OH
S-adenosylmethionine
(SAM)
fig 56b
17.4C: Single-carbon transfer with formyl-THF
There are two important f-THF-dependent formylation steps in the biosynthetic pathways
for purine nucleophiles. Both are simply transamidation reactions: in other words,
conversion by the nucleophilic acyl substitution mechanism (chapter 11) of one amide to
another.
Glycinamide ribonucleotide transformylase reaction:
392
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
O
PO
O
H2N
NH2
HN
+
H
OH OH
O
H
N
O
N
H
H
N
f-THF
O
PO
N
HN
H
N
H2N
C
H
N
R
C
O
H
N
N
O
+
H
N
N
H
O
N
H
OH OH
R
THF
fig 57
Aminoimidazole carboxamide transformylase reaction:
O
O
NH2
N
PO
O
N
NH2
NH2
N
f-THF
THF
PO
O
NH
N
O
OH OH
C H
OH OH
fig 58
Exercise 17.10: Which of the two transformylase reactions above would you expect to be
more thermodynamically favorable? Explain your reasoning.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
393
Chapter 17: Coenzymes
Exercise 17.11: Propose a mechanism for the reaction below, from the histidine
degradation pathway.
H
HN C
H2N
NH
+
O2C
THF
H
N
N
N
H
CO2
N
O
H
C
HN
R
NH
5-formimino THF
fig 59
17.4D: Single-carbon transfer with methylene-THF
CH2-THF serves as a single-carbon donor in a somewhat complicated reaction in the
biosynthesis of the DNA monomer doexythymidine monophosphate (dTMP).
(R = ribose-5-phosphate)
R
R
N
O
CH2-THF
DHF
N
O
HN
HN
CH3
O
O
dUMP
dTMP
fig 60
The reactions begins with the five-membered ring of CH2-THF breaking apart to create
an imine intermediate (step 1 below). The imine becomes an electrophile in a conjugate
addition step (section 13.4, steps 2-3 below). Note that after step 1, the second ring in the
pterin system of the coenzyme is abbreviated for clarity.
394
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
H
H2N
N
H
N
N
N
H
N
O
H
CH2-THF
H
step 1
C
N
N
H A
O
:B
H
S
R
CH2 HN R
H
S
HN
HN
N
N
O
step 3
O
HN
O
N
CH2 HN R
R
H
step 2
O
enz
N
S
R
N
O
R
enz
enz
dUMP
fig 61
Next, tetrahydrofolate is eliminated in an E1cb elimination mechanism (steps 4 and 5).
Notice that this is where the single carbon is transferred from methyl-THF to dUMP.
H
N
HN
O
A
N
R
CH2 HN R
H
:B
S
N
H
N
O
CH2 HN R
step 4
HN
O
enz
N
H
N
O
N
H
H
N
R
HN
THF
R
step 5
O
S
enz
CH2
HN
O
N
R
S
enz
fig 62
The final step in the mechanism is where it gets really interesting: a hydride ion is
transferred from the tetrahydrofolate coenzyme to the methylene (CH2) group on the
deoxynucleotide substrate. Essentially, this is a conjugated SN2 step with hydride as the
nucleophile and the active site cysteine as the leaving group.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
395
Chapter 17: Coenzymes
H
N
H
N
N
DHF
N
H
H
B:
O
O
step 6
HN R
O
CH2
HN
N
R
S
H A
enz
HN R
H
C H
HN
O
N
R
dTMP
fig 63
396
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
HS
enz
Chapter 17: Coenzymes
Key learning objectives for this chapter
After completing this chapter, you should be able to:
Understand how pyridoxal phosphate (PLP) acts as an 'electron sink' in a variety of
reactions in amino acid metabolism.
Recognize and draw mechanisms for PLP-dependent transformations of the following
types:
racemization
decarboxylation
transamination
retroaldol cleavage
retro-Claisen cleavage
β-elimination
β-substitution
γ-elimination
γ-substitution
Recognize transformations - amino acid decarboxylation and transamination, for
example - in which chemical steps occur that simply don't 'make sense' unless the
electron sink role of PLP is taken into account.
Understand how the orientation of the substrate in relation to the plane formed by the
conjugated π system of PLP is a major factor in catalysis of PLP-dependent reactions.
Understand how thiamine diphosphate (ThDP) acts as an 'electron sink' in a variety of
reactions in which a bond to a carbonyl carbon is broken, and how these steps do not
'make sense' unless the electron sink role of ThDP is taken into account.
Recognize transformations for which ThDP is likely required, and be able to draw
reasonable mechanisms for them.
Understand how ThDP acts in tandem with lipoamide, flavin, and nicotinamide in the
reaction catalyzed by pyruvate dehydrogenase.
Recognize folate in its various forms - DHF, THF, f-THF, CH2-THF, and CH3-THF functions in a variety of one-carbon transfer steps. Be able to recognize the oxidation
state of the carbon being transferred in a folate-depenent step.
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
397
Chapter 17: Coenzymes
398
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
Problems
P17.1: Here is a chance to test your ability to recognize reactions catalyzed by enzymes
using three coenzymes - thiamine diphosphate, pyridoxal phosphate, and folate - that we
have studied in this chapter. For each generalized reaction, look carefully at the
transformation that is taking place, and decide which of the three coenzymes is likely to
be required. Then, draw the single mechanistic step by which the bond identified by an
arrow is broken or formed. In the cases where a double bond is indicated, show the
mechanistic step in which the σ bond is formed. In each case, your drawing should
include the structure of the reactive part of the coenzyme, and should clearly show the
role it plays in catalyzing the mechanistic step you are drawing.
a)
CO2
O
R
O
H
R
CO2
b)
CO2
R
H3N
R
H3N
CO2
c)
O
O
R2
R1
H
R1
+
H
R2
O
OH
d)
SR2
H3N
R1
R2SH
H3N
R1
e)
O
R
NH2
R
N
H
H
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
399
Chapter 17: Coenzymes
f)
R3SH
O
O
R2
R1
R1
R2
R3S
+
H
O
O
g)
R3SH
SR2
H3N
R2SH
SR3
H3N
R1
R1
h)
SR2
H3N
R2SH
R1
H3N
R1
i)
R3SH
SR2
H3N
R2SH
SR3
H3N
R1
R1
j)
SR2
H3N
R3SH
SR3
R2SH
R1
H3N
R1
k)
R1
CO2
R3SH
O
O
CO2
+
R2
SR3
O
R2
R1
O
l)
R2
H3N
R2
O
R1
+
R3
R4
O
NH3
R1
+
R3
m)
400
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
R4
Chapter 17: Coenzymes
O
R1
O
R2
R2
R1
P17.2: The final step in the biosynthesis of the amino acid is a PLP-dependent
condensation between serine and indole, shown below. The reaction mechanism involves
steps that are familiar from this chapter, but also incorporates a reaction type we studied
in chapter 14. Propose a mechanism.
H
N
H
N
OH
H2O
+
H3N
PLP
CO2
serine
H3N
CO2
tryptophan
indole
P17.3: Draw a reasonable mechanism for the following reaction, identifying the species
denoted by questions marks. Biochemistry 2012, 51, 3059
OH
OH
HN
N
?
?
O
H3N
CO2
H3N
CO2
P17.4: Propose a mechanism for the reaction below, which is part of the anaerobic
catabolism of alcohols in some species of bacteria. ChemBioChem 2014, 15, 389
H2O
O
O
CO2
[coenzyme]
O
P17.5: Identify cosubstrate A and propose a mechanism for the reaction shown below,
which was reported to occur in the thermophilic bacterium Thermosporothrix hazakensis.
ChemBioChem 2014 15, 527
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
401
Chapter 17: Coenzymes
2CO2
O
A
+
CO2
OH
[coenzyme]
O
P17.6: Propose a mechanism for each of the reactions below, being sure to show the role
played by the coenzyme (you need to determine which coenzyme is needed in each case).
a) Biomed. Res. Int. 2013 E-pub http://www.ncbi.nlm.nih.gov/pubmed/24175284
OH
O
+
H3N
OH
CO2
HSCoA
R
CO2
[coenzyme]
SCoA
R
H3N
O
palmitoyl-CoA
serine
b) (E.C. 2.2.1.6)
CO2
O
2
H3C
HO CH3
H3C
CO2
[coenzyme]
CO2
O
pyruvate
(S)-acetolactate
c) Angew. Chem. Int. Ed. Engl. 2014, 53, 1943.
http://onlinelibrary.wiley.com/doi/10.1002/anie.201307989/full Fig 1 Orf R rxn
H2N
NH
H2O
HN
HO
H3N
H
N
HN
HN
H OH
CO2
OH
[coenzyme]
H3N
CO2
P17.7: Acetohydroxybutyrate is formed in a coenzyme-dependent reaction between
pyruvate and a 4-carbon compound. What is a likely second substrate, coenzyme, and
by-product (indicated below with a question mark)?
http://www.sciencedirect.com/science/article/pii/S1367593105001043
402
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
O
?
O
H3C
CO2
+ a 4-carbon substrate
H3C
[coenzye]
HO
CO2
acetohydroxybutyrate
pyruvate
P17.8:
a) The 'benzoin condensation' reaction was discovered in the 19th century, and led
eventually to a better understanding of ThDP-dependent reactions in the cell. In a
traditional benzoin condensation reaction, cyanide ion (instead of ThDP) plays the role of
electron acceptor. Enzyme-catalyzed benzoin condensation reactions are also known to
occur in some bacteria: Pseudomonas fluorescens, for example, contains an enzyme that
catalyzes the synthesis of (R)-benzoin.
O
O
H
2
OH
benzoin
benzaldehyde
a) Draw a mechanism for the enzyme-catalyzed benzoin condensation reaction.
b) Draw a mechanism for the cyanide-catalyzed benzoin condensation reaction (nonenzymatic, basic conditions).
c) The following ThDP-reaction was recently reported to be part of the biosynthetic
pathway for clavulanic acid, a compound that inhibits the action of β-lactamases (βlactamases are bacterial enzymes that hydrolyze penicillin-based antibiotic drugs,
rendering them ineffective). As is typical for ThDP-dependent reactions, the first step is
addition of the ylide form of the coenzyme to the substrate carbonyl. The next steps are
(in order): dehydration, tautomerization, elimination of phosphate, conjugate addition of
arginine, and finally hydrolytic cleavage of the coenzyme-product bond. Draw out a
complete mechanism that corresponds to this description. J. Am. Chem. Soc. 2007, 129,
15750.
OH
arginine
H2O
Pi
O
NH
O
O
OP
HN
[ThDP]
CO2
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
N
H
NH2
403
Chapter 17: Coenzymes
P17.9: Propose a mechanism for the reaction catalyzed by aspartate β-decarboxylase,
which converts aspartate to alanine in a PLP-dependent reaction.
J. Mol. Biol. 2009, 388, 98.
P17.10: As we saw in this chapter, PLP-dependent enzymes usually catalyze reactions
involving amino acid substrates. Here is an exception, a PLP-dependent β-elimination
reaction in the folate biosynthetic pathway. Propose a mechanism for this reaction.
J. Biol. Chem 2013, 288, 22985
CO2
CO2
pyruvate
CH2
O
CO2
NH2
NH2
chorismate
para-aminobenzoate
Pathway prediction:
P17.11: Propose a multistep pathway for each of the following transformations. All
involve at least one step requiring PLP, ThDP, or folate.
a) Below is portion of the biosynthesis of a modified membrane lipid in Salmonella and
other pathogenic bacteria. The modified membrane confers antibiotic resistance to the
bacterium. Biochemistry 2014, 53, 796
OH
HO
HO
NH2
O
O
H
OH
H
HO
OH
OH
O
UDP
b) Below is the biosynthetic pathway for phenethanol in yeast. Phenethanol, which has a
rose scent, is commonly used as a fragrance - this pathway has been proposed as a
potential 'green' enzymatic synthesis to replace the traditional industrial synthesis, which
uses toxic reagents. J. Appl. Microbiol. 2009, 106, 534.
404
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Chapter 17: Coenzymes
H3N
CO2
OH
phenethanol
phenylalanine
c) Below is an incomplete pathway diagram for the biosynthesis of the amino acid lysine,
starting from aspartate. Fill in the missing steps and reactants/coenzymes to complete the
diagram. The solid dot and dashed circle are provided to help you to trace two of the
carbons from substrate to product.
O
OH
O
O
O2C
H3N
CO2
N
O2C
CO2
CO2
HN
aspartate
O
CO2
H3N
H3N
CO2
lysine
d) Below is the second half of the tryptophan degradation pathway. Fill in the missing
steps and reactants/coenzymes to complete the diagram.
O
CO2
O
O
NH3
O2C
SCoA
2
H3C
SCoA
acetyl CoA
d) Below is an incomplete pathway diagram for the biosynthesis of inosine
monophosphate, a precursor to the nucleotides adenosine monophosphate (AMP) and
guanosine monophosphate (GMP). Fill in missing steps and reactants/coenzymes to
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
405
Chapter 17: Coenzymes
complete the diagram. Note that one enzymatic step is provided (this is a carboxylation
reaction of a type that we have not studied).
NH2
=
PO
ribose
phosphate
PO
O
O
O
NH2
OH
these two carbons come
from glycine
H
N
HN
OH OH
OH OH
NH
ribose
phosphate
ribose phosphate
nitrogen comes
from aspartate
O
ATP
HCO3
N
N
ribose
phosphate
NH2
ADP
Pi
CO2
N
N
NH
N
NH2
N
ribose
phosphate
N
ribose
phosphate
inosine
monophosphate
406
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Tables
List of tables:
Table 1: Some characteristic absorption frequencies in IR spectroscopy
Table 2: Typical values for 1H-NMR chemical shifts
Table 3: Typical values for 13C-NMR chemical shifts
Table 4: Typical coupling constants in NMR
Table 5: The 20 common amino acids
Table 6: Structures of common coenzymes
Table 7: Representative acid constants
Table 8: Some common laboratory solvents, acids, and bases
Table 9: Examples of common functional groups in organic chemistry
Organic Chemistry with a Biological Emphasis
Tim Soderberg
407
Table 1: Some characteristic absorption frequencies in IR spectroscopy
Bond type
frequency (cm-1)
C-H alkanes
2950 – 2850
C-H alkenes
3080 – 3020
C-H aldehyde
~2900
C-H alkyne
~3300
alkyne triple bond
2250 – 2100 (s)
alkene double bond
1680 - 1620(s)
carbonyl, ketone
1725 – 1700 (s)
carbonyl, aldehyde
1740 – 1720 (s)
carbonyl, ester
1750 – 1730 (s)
carbonyl, acid
1725 – 1700 (s)
carbonyl, amide
1690 – 1650 (s)
O-H, alcohols
3600 – 3200 (s, broad)
O-H, acids
3000 – 2500 (broad)
C-O, alcohols, esters, ethers
1300 - 1000
s = strong absorbance
408
Organic Chemistry with a Biological Emphasis
Tim Soderberg
Table 2: Typical values for 1H-NMR chemical shifts
Hydrogen type
Chemical shift (ppm)
RCH3
0.9 - 1.0
RCH2R
1.2 - 1.7
R3CH
1.5 – 2.0
O
C
R
CH3
R
CH3
C C
R
1.5 – 1.8
R
R
2.0 – 2.3
RNH2
1-3
ArCH3
2.2 – 2.4
C
C
O
R
H
CH3
2.3 – 3.0
3.7 – 3.9
O
C
R
CH3
O
3.7 – 3.9
1-5
ROH
R
H
C C
R
R
3.7 – 6.5
O
R
C
N
R
5-9
H
6.0 – 8.7
ArH
O
R
C
H
9.5 – 10.0
O
R
C
OH
10 - 13
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
Organic Chemistry with a Biological Emphasis
Tim Soderberg
409
Table 3: Typical values for 13C-NMR chemical shifts
Carbon type
Chemical shift (ppm)
RCH3
13 - 16
RCH2R
16 - 25
R3CH
25 - 35
O
R
C
O
CH3
18 - 22
O
C
R
CH3
28 - 32
RCH2NHR
35 - 45
RCH2OH
50 - 65
R
C
C
R
ROCH2R
65 - 70
50 - 75
O
R
O
CH2R
R
50 - 75
H
C C
R
H
R
115 - 120
H
C C
R
H
aromatic carbon
125 - 140
125 - 150
O
C
R
X
(carboxylic acid derivatives)
165 - 185
O
R
C
H
190 - 200
O
R
410
C
R
200 - 220
Organic Chemistry with a Biological Emphasis
Tim Soderberg
Table 4: Typical coupling constants in NMR
H-H coupling
J (Hz)
C-H coupling
R
R R
R
C C R
6 -8
R
R O
C C H
R
2-3
H
H
125 - 130
C H
R
H H
R
J (Hz)
H
C C
R
150 - 170
R
R
C C
R
H
R
12 - 18
R
C C
H
H
R
6 - 12
H
C C
R
H
0-2
H
H
6 - 10
H
1- 3
H
Organic Chemistry with a Biological Emphasis
Tim Soderberg
411
Table 5: The 20 common amino acids
H
H3N
CH3
H
H3N
CO2
H
CO2
Glycine
(Gly, G)
H
H3N
Alanine
(Ala, A)
CO2
Valine
(Val, V)
H3N
Leucine
(Leu, L)
N
H3N
CO2
CO2
Tyrosine
(Tyr, Y)
HO
CO2
Tryptophan
(Trp, W)
CH3
H
CO2
Serine
(Ser, S)
H3N
H 3N
Threonine
(Thr, T)
CO2
O
Proline
(Pro, P)
412
H
H 3N
H3N
Arginine
(Arg, R)
O
CO2
Glutamate
(Glu, E)
H
H3N
H
CO2
H3N
Methionine
(Met, M)
H
CO2
H
CO2
H3N
NH
N
H
H3N
H
CO2
Lysine
(Lys, K)
H 2N
H 3N
Aspartate
(Asp, D)
Organic Chemistry with a Biological Emphasis
Tim Soderberg
CO2
Histidine
(His, H)
O
O
O
CO2
CO2
Cysteine
(Cys, C)
H
N
NH2
H
O
N
H2
Isoleucine
(Ile, I)
H
H
H 2N
H3N
CO2
SH
H3N
OH
H3N
S
H
H
Phenylalanine
(Phe, F)
CO2
H
HO
H3N
H
H
H
H
H3N
CO2
Glutamine
(Gln, Q)
H3N
NH2
CO2
Asparagine
(Asn, N)
Table 6: Structures of common coenzymes
NH2
N
O
HS
O
N
N
H
H
O
O P O P O
O
OH
N
O P O P O P O
O
O
O
N
H3N
S
N
OH
OH
S-adenosyl methionine
(SAM)
O
O
O P O
H
OH
S
O
N
NH2
N
N
PPO
CH3
N
CH3
H
CH3
thiamine diphosphate
TPP)
pyridoxal phosphate
(PLP)
NH3
H
N
O2C
lipoate
N
thiazole ring
H
S
N
O
HO
adenosine triphosphate
(ATP)
CO2
N
H3C
O
HO
S
NH2
CO2
N
N
O
OH
O P O
NH2
O
N
O
Coenzyme A
(HSCoA)
O
N
O
O
O
N
O
O
O
H
N
CO2
H
O
N H
N
H
H
HS
S
Glutathione
(GSH)
biotin
CO2
Table 5
Organic Chemistry with a Biological Emphasis
Tim Soderberg
413
O
H
NH2
NH2
O
N
O
O P O P
O
nicotinamide group
HO
N
O
O
O
N
N
N
O
OH
HO
OH
nicotinanide adenine dinucleotide - oxidized form
(NAD+, or NADP+ if phosphorylated at arrow position)
O
H
NH2
H
NH2
O
N
O
O P
O
HO
N
O
O P O
O
N
N
N
O
OH
HO
OH
nicotinanide adenine dinucleotide - reduced form
(NADH, or NADPH if phosphorylated at arrow position)
O
CO2
N
H
O
H
H2N
H
N
N
N
CO2
H
N
O
H
H
N
H2N
414
H
R
C N
N
N
H
N
N
H
H
tetrahydrofolate (THF)
H
5,10-methylenetetrahydrofolate
Organic Chemistry with a Biological Emphasis
Tim Soderberg
O
N
N
N
N
H
flavin group
O
N
O
CH2
OH
O
HO
N
O
H 2C O P O P O
O
O
N
O
HO
OH
N
O
HO
H 2C O P O
OH
O
flavin mononucleotide, oxidized form
(FMN)
flavin adenine dinucleotide, oxidized form
(FAD)
O
N
N
N
N
CH2
H
H
H
NH2
OH
O
N
O
H2C O P O P O
O
O
N
O
HO
HO
N
H
CH2
N
O
HO
H
N
NH2
HO
N
O
N
N
N
N
N
CH2
H
H
O
HO
OH
N
HO
O
O
H2C O P O
HO
OH
flavin adenine dinucleotide, reduced form
(FADH2)
O
flavin mononucleotide,reduced form
(FMNH2)
Organic Chemistry with a Biological Emphasis
Tim Soderberg
415
Table 7: Representative acid constants.
acid
pKa
conjugate base
O
O
HO
S OH
O
-10
HO
S O
O
sulfuric acid
HCl
H 3O
+
-
-7
Cl
-1.7
H 2O
O
O
N
O
OH
nitric acid
-1.4
O
N
O
O
O
R O P OH
(i)
1.0
R O P
O
OH
OH
O
O
HO P OH
OH
(ii)
2.2
HO P
O
OH
phosphoric acid
3.2
HF
F
NH3
-
NH2
4.6
O
R
C
O
OH
4-5
N H
R
C
O
N
(ii)
pyridinium
416
5.3
Organic Chemistry with a Biological Emphasis
Tim Soderberg
pyridine
acid
pKa
conjugate base
O
O
C
HO
OH
HO
6.4
C
O
carbonic acid
bicarbonate
O
O
R O P
O
OH
6.5(i)
R
O P
O
O
O
HO P
O
(ii)
7.2
O P O
OH
OH
O H O
H3C C C C CH3
O
9.0
O
H3C C C C CH3
H
H
9.2
HCN
CN
-
NH3
NH4
ammonium
OH
9.2
ammonia
9.9(ii)
O
phenol
phenolate
O
O
HO
C
O
10.3(ii)
bicarbonate
RSH
RNH3
+
O P O
O
C
O
carbonate
-
10-11
RS
10 -11
RNH2
O
O
(i)
12.3
O P
O
O
OH
H 2O
O
15.7
Organic Chemistry with a Biological Emphasis
Tim Soderberg
OH
-
417
acid
pKa
conjugate base
O
O
R
C
NH2
RCH2OH
17
R
16
C
NH
RCH2O
-
O
O
R
C
C
R
H H
19-20
R
C
C
R
H
RCCH
terminal alkyne
25
RCC
H2
35
H
NH3
ammonia
38
NH2
-
-
-
All pKa values, unless otherwise noted, are taken from March, Jerry, Advanced Organic Chemistry, Fourth
Edition, Wiley, New York, 1992.
(i)
Silva, J.J.R. Fraústo da, The Biological Chemistry of the Elements: the Inorganic Chemistry of Life, 2nd
Edition, Oxford, New York, 2001.
(ii)
Lide, David R. (ed.) The CRC Handbook of Chemistry and Physics, CRC Press, Boca Raton, FL, 1995.
418
Organic Chemistry with a Biological Emphasis
Tim Soderberg
Table 8: Some common laboratory solvents, acids, and bases (fig 10)
Solvents
CH3CH2OH
CH3OH
H2O
ethanol
methanol
water
O
O
H3C
O
C
CH3
acetone
S
H
N C CH3
H3C
CH3
acetonitrile
dimethyl sulfoxide
(DMSO)
C
N
CH3
CH3
dimethylformamide
CH3
O
O
diethylether
tetrahydrofuran
(THF)
H
H
Cl C H
Cl C Cl
Cl
Cl
dichloromethane
(methylene chloride)
chloroform
benzene
toluene
Cl
Cl C Cl
Cl
carbon tetrachloride
Acids
O
O
HO
S OH
O
sulfuric acid
HO
S
O
p-toluenesulfonic acid
(TsOH)
O
CH3
HO P OH
OH
phosphoric acid
Organic Chemistry with a Biological Emphasis
Tim Soderberg
HCl
hydrochloric acid
419
Bases
very strong bases:
Li
N
Li
lithium diisopropylamide
(LDA)
CH2CH2CH2CH3
N-butyllithium
NaH
Na NH2
sodium hydride
sodium amide
weaker bases:
CH2CH3
CH3
N CH2CH3
H3C C O K
CH3
N
CH2CH3
potassium tert-butoxide
pyridine
triethylamine
O
Na O
C
O
OH
sodium bicarbonate
NaHCO3
420
K O
C
O K
potasium carbonate
K2CO3
NaOH
sodium hydroxide
Organic Chemistry with a Biological Emphasis
Tim Soderberg
Table 9: Examples of common functional groups in organic chemistry
H
H
H C
C
H
H
alkane
alkene
H
alkyne
H
H
C C
H
O
ketone
H3C
H C C H
H3C
imine
(Schiff base)
alkyl halide
H3C
H
alcohol
H
H3C C
H3C
H
H3C
O
H3C
S
CH3
C
O
O P
O
O
CH3
O
acid chloride
H 3C
sulfide
N
O
acyl phosphate
H3C C NH2
ether
CH3
H
H
amine
S
H 3C
H
CH3
O
amide
SH
OH
O
H 3C
H
CH3
O
thioester
H3C C OH
thiol
C
H3C
H
CH3
O
ester
H C Cl
C
H
O
carboxylic acid
H
C
N
H3C
aromatic
hydrocarbon
CH3
O
aldehyde
H
C
CH3
phosphate ester
C
Cl
O
O P OCH3
O
phenol
OH
phosphate
diester
O
O P OCH3
OCH3
Organic Chemistry with a Biological Emphasis
Tim Soderberg
421
Appendix: EC Numbers
Appendix: Enzymatic reactions by
metabolic pathway and EC number
The EC (European Commission) number is a classification system for enzymes,
organized by the type of reaction catalyzed. You can use the list below to search this
textbook for information about the reaction catalyzed by a given enzyme. You can also
use the EC numbers to search for information in databases such as SwissProt Enzyme
Nomenclature Database (http://enzyme.expasy.org/) or the BRENDA Comprehensive
Enzyme Information System (www.brenda-enzymes.org/).
Glycolysis
Hexose kinase (EC 2.7.1.1)
Phosphoglucose isomerase (EC 5.3.1.9)
Phosphofructokinase (EC 2.7.1.56)
Fructose 1,6-bisphosphate aldolase (EC 4.1.2.13)
Triose phosphate isomerase (5.3.1.1)
Gyceraldehyde 3-phosphate dehydrogenase (EC 1.2.1.12)
Phosphoglycerate kinase (EC 2.7.2.3)
Phosphoglycerate mutase (EC 5.4.2.1)
Enolase (EC 4.2.1.11)
Pyruvate kinase (EC 2.7.1.40)
2-Keto-3-deoxy-6-phosphogluconate (KDPG) aldolase (EC 4.1.2.14)
Gluconeogenesis
Pyruvate carboxylase (EC 6.4.1.1)
Phosphoenolpyruvate carboxykinase (EC 4.1.1.32)
Phosphoglycerate kinase (EC 2.7.2.3)
Fructose 1,6-bisphosphatase (EC 3.1.3.11)
Glucose 6-phosphatase (EC 3.1.3.9)
422
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
Citric Acid (TCA) Cycle
Pyruvate dehydrogenase complex (EC 1.2.4.1)
Citrate synthase (EC 2.3.3.8)
Aconitase (EC 4.2.1.3)
Isocitrate dehydrogenase (EC 1.1.1.42)
α-ketoglutarate dehydrogenase complex:
Oxoglutarate dehydrogenase (EC 1.2.4.2)
Dihydrolipoyl succinyltransferase (EC 2.3.1.61)
Dihydrolipoyl dehydrogenase (EC 1.8.1.4)
Succinyl CoA synthetase (EC 6.2.1.4)
Succinate dehydrogenase (EC 1.3.5.1)
Fumarase (EC 4.2.1.2)
Malate dehydrogenase (EC 1.1.1.37)
Fermentation
Pyruvate decarboxylase (EC 4.1.1.1)
Alcohol dehydrogenase (EC 1.1.1.1)
Lactate dehydrogenase ((EC 1.1.1.27)
Pentose phosphate pathway/Calvin Cycle
Glucose 6-phosphate dehydrogenase (EC 1.1.1.49)
Gluconolactonase (EC 3.1.1.17)
6-Phosphogluconate dehydrogenase (EC 1.1.1.43)
Phosphopentose epimerase (EC 5.1.3.1)
Phosphopentose isomerase (EC 5.3.1.6)
Transketolase (EC 2.2.1.1)
Transaldolase (EC 2.2.1.2)
Rubisco (EC 4.1.1.39)
Aldolase (EC 4.1.2.13)
Fatty acid oxidation
Acyl CoA synthetase (EC 6.2.1.1)
Carnitine acyltransferase (EC 2.3.1.21)
Acyl CoA dehydrogenase (eg. EC 1.3.99.13)
Enoyl CoA hydratase (eg. 4.2.1.74)
3-hydroxyacyl CoA dehydrogenase (EC 1.1.1.35)
β-Keto thiolase (EC 2.3.1.16)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
423
Appendix: EC Numbers
Unsaturated fatty acids:
cis-Enoyl-CoA isomerase (eg. EC 5.3.3.8)
2,4-Dienoyl CoA reductase (EC 1.3.1.34)
Glycerol phosphate dehydrogenase (EC 1.1.1.8)
Fatty acid biosynthesis
Acetyl CoA carboxylase (EC 6.4.1.2)
Acyl CoA synthetase (EC 6.2.1.1)
ACP transacylase (EC 2.3.1.38)
β-ketoacyl-ACP synthase (EC 2.3.1.41)
β-ketoacyl-ACP hydrogenase (EC 1.1.1.35)
3-hydroxyacyl dehydratase (EC 4.2.1.58)
Enoyl-ACP reductase (EC 1.3.1.10)
Acyl-CoA dehydrogenase (EC 1.3.99.3)
Monoacylglycerol acyltransferase (EC 2.3.1.22)
Isoprenoid biosynthesis
Mevalonate pathway (from acetyl CoA):
Acetoacetyl CoA acetyltransferase (EC 2.3.1.9)
3-hydroxy-3-methylglutaryl-CoA synthase (EC 2.3.3.10)
HMG-CoA reductase (EC 1.1.1.34)
Mevalonate kinase (EC 2.7.1.36)
Phosphomevalonate kinase (EC 2.7.4.2)
Mevalonate diphosphate decarboxylase (EC 4.1.1.33)
Deoxyxylulose pathway (from pyruvate and glyceraldehyde phosphate):
Deoxyxylulose phosphate synthase (EC 2.2.1.7)
Deoxyxylulose phosphate reductoisomerase (EC 1.1.1.267
MEP cytidylyltransferase (EC 2.7.7.60)
CDP-ME kinase (EC 2.7.1.148)
Methylerithritol 2,4-cyclodiphosphate synthase (EC 4.6.1.12)
424
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
other:
isopentenyl diphosphate isomerase (EC 5.3.3.2)
geranyl diphosphate synthase (EC 2.5.1.1)
farnesyl diphosphate synthase (EC 2.5.1.10)
squalene synthase (EC 2.5.1.21)
oxidosqualene cyclase (EC 5.4.99.7)
DMAPP-tryptophan synthase (EC 2.5.1.34)
Deoxyribonucleotide biosynthesis
PRPP synthetase (EC 2.7.6.1)
UMP (from ammonia, bicarbonate, and aspartate):
Carbamoyl phosphate synthase (EC 6.3.5.5)
Aspartate carbamoyltransferase (EC 2.1.3.2)
Dihydroorotase (EC 2.5.2.3) 4
Dihydroorotate dehydrogenase (EC 1.3.3.1)
Orotate phosphoribosyltransferase (EC 2.4.2.10)
Orotidine monophosphate decarboxylase (EC 4.1.1.23)
TMP (from dUMP):
Thymidylate synthase (EC 2.1.1.45)
CTP: (from UTP):
CTP synthase (EC 6.3.4.2)
IMP (from PRPP):
Glutamine phosphoribosyl amidotransferase (EC 2.4.2.14)
Glycinamide ribonucleotide synthetase (EC 6.3.4.13)
GAR transformylase (EC 2.1.2.2)
FGAM synthetase (EC 6.3.5.3)
Aminoimidazole ribonucleotide synthetase (EC 6.3.3.1)
Aminoimidazole ribonucleotide carboxylase (EC 4.2.1.1)
SAICAR synthetase (EC 6.3.2.6)
Adenylosuccinate lyase (EC 4.3.2.2)
AICAR transformylase (EC 2.1.2.3)
IMP cyclohydrolase (EC 3.5.4.10)
AMP (from IMP):
Adenylosuccinate synthetase (EC 6.3.4.4)
Adenylosuccinate lyase (EC 4.3.2.2)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
425
Appendix: EC Numbers
GMP (from IMP):
IMP dehydrogenase (EC 1.1.1.205)
GMP synthetase (EC 6.3.5.2)
Deoxyribonucleotides:
Ribonucleotide reductase (EC 1.17.4.1)
Nucleotide degradation
Cytidine (to uridine):
Cytidine deaminase (EC 3.5.4.5)
Uridine (to malonyl CoA):
Uridine phosphorylase (EC 2.4.2.3)
Dihydropyrimidine dehydrogenase (EC 1.3.1.2)
Dihydropyrimidase (EC 3.5.2.2)
β-ureidopropionase (EC 3.5.1.6)
Thymidine (to succinyl CoA):
Thymidine phosphorylase (EC 2.4.2.4)
Dihydropyrimidine dehydrogenase (EC 1.3.1.2)
Dihydropyrimidase (EC 3.5.2.2)
β-ureidopropionase (EC 3.5.1.6)
Adenosine (to uric acid):
Adenosine deaminase (EC 3.5.4.4)
Purine nucleoside phosphorylase (EC 2.4.2.1)
Xanthine oxidase (EC 1.17.1.4; EC 1.17.3.2)
Guanosine (to uric acid):
Purine nucleoside phosphorylase (EC 2.4.2.1)
Guanine deaminase (EC 3.5.4.3)
Xanthine oxidase (EC 1.17.1.4; EC 1.17.3.2)
426
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
Amino acid biosynthesis
Alanine (from pyruvate):
Alanine transaminase (EC 2.6.1.2)
Alanine racemase (PLP-dependent) (EC 5.1.1.1)
Aspartate (from oxaloacetate):
Aspartate transaminase (EC 2.6.1.1)
Glutamate (from α-ketoglutarate):
Glutamate transaminase (EC 2.6.1.1)
Glutamine (from glutamate):
Glutamine synthase (EC 6.3.1.2)
Asparagine (from aspartate):
Asparagine synthase (EC 6.3.5.4)
Arginine (from glutamate via ornithine):
N-acetylglutamate synthase (EC 2.3.1.1)
Acetylglutamate kinase (EC 2.7.2.8)
N-acetyl-γ-glutamyl-phosphate reductase (1.2.1.38)
Acetylornithine transaminase (EC 2.6.1.11)
Acetylornithine deacetylase (EC 3.5.1.16)
then ornithine to arginine via urea cycle
Proline (from glutamate):
Glutamate-5-kinase (EC 2.7.2.11)
Glutamate-5-semialdehyde dehydrogenase (EC 1.2.1.41)
Pyrroline-5-carboxylate reductase (EC 1.5.1.2)
Serine (from 3-phosphoglycerate):
Phosphoglycerate dehydrogenase (EC 1.1.1.95)
Phosphoserine transaminase (EC 2.6.1.52)
Phosphoserine phosphatase (EC 3.1.3.3)
Cysteine (from serine):
Cystathionine β-­‐synthase (EC 4.2.1.22) Cystathionine γ-lyase (EC 4.4.1.1)
. . . or (in bacteria) :
O-acetylserine sulfhydrolase (EC 2.5.1.47)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
427
Appendix: EC Numbers
Glycine (from serine):
Serine hydroxymethyltransferase (EC 2.1.2.1)
Lysine (from aspartate):
Aspartate kinase (EC 2.7.2.4) P16.3
Aspartate-semialdehyde dehydrogenase (EC 1.2.1.11)
Dihydrodipicolinate synthase (EC 4.2.1.52)
Dihydrodipicolinate reductase (EC 1.3.1.26)
Tetrahydropyridine-2-carboxylate N-succinyltransferase (EC 2.3.1.117)
Succinyl-diaminopimelate transaminase (EC 2.6.1.17)
Succinyl-diaminopimelate desuccinylase (EC 3.5.1.18)
Diaminopimelate decarboxylase (EC 4.1.1.20)
Methionine (from aspartate via homoserine): Aspartate kinase (EC 2.7.2.4)
Aspartate-semialdehyde dehydrogenase (EC 1.2.1.11)
Homoserine dehydrogenase (EC 1.1.1.3)
Homoserine transsuccinylase (EC 2.3.1.46)
Cystathionine g-synthase (EC 4.2.1.22)
Cystathionine β-lyase (EC 4.4.1.8)
Methionine synthase (EC 2.1.1.13)
Threonine (from homoserine):
Homoserine kinase (EC 2.7.1.39)
Threonine synthase (EC 4.2.3.1)
Isoleucine, leucine, valine (from pyruvate):
Acetolactate synthase (EC 2.2.1.6)
Ketol acid reductoisomerase (EC 1.1.1.86)
Dihydroxyacid dehydratase (EC 4.2.1.9)
Isoleucine:
Branched-chain-amino-acid transaminase (EC 2.6.1.42)
Valine:
Valine—pyruvate transaminase (EC 2.6.1.66)
Leucine:
2-isopropylmalate synthase (EC 2.3.3.13)
Isopropylmalate isomerase (EC 4.2.1.33)
3-isopropylmalate dehydrogenase (EC 1.1.1.85
Leucine transaminase (EC 2.6.1.6)
428
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
Aromatic amino acids
Erythrose-4P to chorismate:
DAHP synthase (EC 2.5.1.54)
Dehydroquinate synthase (EC 4.2.3.4)
Dehydroquinate dehydratase (EC 4.2.1.10)
Shikimate dehydrogenase (EC 1.1.1.25)
shikimate kinase (EC 2.7.1.71) P10.2
5-enolpyruvylshikimate-3-phosphate (EPSP) synthase (EC 2.5.1.19)
Chorismate synthase (EC 4.2.3.5)
Chorismate to tryptophan:
Anthranilate synthase (EC 4.1.3.27)
Anthranilate phosphoribosyltransferase (EC 2.4.2.18)
Phosphoribosyl anthranilate isomerase (EC 5.3.1.24)
Indole-3-glycerol phosphate synthase (EC 4.1.1.48)
Tryptophan synthase (EC 4.2.1.20)
Chorismate to phenylalanine/tyrosine
Chorismate mutase (EC 5.4.99.5)
Prephenate decarboxylase (EC 4.2.1.51)
Aromatic-amino-acid transaminase (EC 2.6.1.57)
Tyrosine transaminase (EC 2.6.1.5)
Histidine (from PRPP and ATP):
ATP phosphoribosyltransferase (EC 2.4.2.17)
Phosphoribosyl-ATP diphosphatase (EC 3..6.1.31)
Phosphoribosyl-AMP cyclohydrolase (EC 3.5.4.19)
1-(5-phosphoribosyl)-5-[(5-phosphoribosylamino)methylideneamino]imidazole-4carboxamide isomerase (EC 5.3.1.16)
Imidazole glycerol-phosphate synthase (EC 4.3.1.B2)
Imidazoleglycerol-phosphate dehydratase (4.2.1.19)
Histidinol-phosphate transaminase (EC 2.6.1.9)
Histidinol-phosphatase (EC 3.1.3.15)
Histidinol dehydrogenase (EC 1.1.1.23)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
429
Appendix: EC Numbers
Amino acid degradation
Transaminase (EC 2.6.1.1, EC 2.6.1.2)
Carbamoyl phosphate synthetase (EC 6.3.4.16)
Urea cycle
Ornithine transcarbamylase (EC 2.1.3.3)
Argininosuccinate synthetase (EC 6.3.4.5) 1D
Arginosuccinate lyase (EC 4.3.2.1)
Arginase (EC 3.5.3.1)
Alanine (to pyruvate and glutamate):
Alanine transaminase (EC 2.6.1.2)
Serine
to pyruvate:
Serine dehydratase (EC 4.3.1.17)
to glycine:
Serine hydroxymethyltransferase (EC 2.1.2.1)
Glycine (glycine cleavage system):
Glycine dehydrogenase (decarboxylating) (EC 1.4.4.2)
Aminomethyltransferase (EC 2.1.2.10)
Cysteine (to pyruvate and SO2):
Cysteine dioxygenase (EC 1.13.11.20)
Aspartate transaminase (2.6.1.1)
Threonine
pathway 1 (to glycine and acetyl CoA):
Threonine dehydrogenase (EC 1.1.1.103)
Glycine C-acetyltransferase (EC 2.3.1.29)
pathway 2 (to glycine and acetaldehyde):
Threonine aldolase (EC 4.1.2.5)
pathway 3 (to succinyl-CoA via propionyl CoA):
Threonine dehydratase (EC 4.3.1.19)
2-oxobutanoate dehydrogenase (EC 1.2.4.4)
Propionyl-CoA carboxylase (EC 6.4.1.3)
Methylmalonyl-CoA mutase (EC 5.4.99.2)
430
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
Tryptophan (to glutaryl-CoA):
Tryptophan 2,3-dioxygenase (EC 1.13.11.11)
Acylformamidase (EC 3.5.1.19)
Kynurenine 3-monooxygenase (EC 1.14.13.9)
Kynurenimase (EC 3.7.1.3)
3-hydroxyanthranilate 3,4-dioxygenase (EC 1.13.11.6)
Aminocarboxymuconate-semialdehyde decarboxylase (EC 4.1.1.45)
2-aminomuconate semialdehyde dehydrogenase (EC 1.2.1.32)
2-aminomuconate deaminase (EC 3.5.99.5)
2-oxoglutarate dehydrogenase (EC 1.2.4.2)
Asparagine (to aspartate):
Asparaginase (EC 3.5.1.1)
Aspartate
to oxaloacetate:
Aspartate transaminase (EC 2.6.1.1)
to fumarate:
Aspartate-ammonia lyase (EC 4.3.1.1)
Glutamine (to glutamate):
Glutaminase (EC 3.5.1.2)
Glutamate (to α-ketoglutarate):
Glutamate dehydrogenase (EC 1.4.1.2)
Arginine (to glutamate):
Arginase (EC 3.5.3.1)
Ornithine transaminase (EC 2.6.1.13)
Glutamate semialdehyde dehydrogenase (EC 1.2.1.41)
Histidine (to glutamate):
Histidine ammonia-lyase (EC 4.3.1.3)
Urocanate hydratase (EC 4.2.1.49)
Imidazolonepropionase (EC 3.5.2.7)
Formimidoylglutamase (EC 3.5.3.8)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
431
Appendix: EC Numbers
Valine, isoleucine, leucine:
Banched chain amino acid transaminase (EC 2.6.1.42)
Branched chain ketoacid dehydrogenase complex (EC 1.2.4.4)
Acyl CoA dehydrogenase (eg. EC 1.3.99.13)
Valine (to succinyl CoA):
Enoyl-CoA hydratase (EC 4.2.1.17)
3-hydroxyisobutyryl-CoA hydrolase (EC 3.1.2.4)
3-hydroxyisobutyrate dehydrogenase (EC 1.1.1.31)
Methylmalonate-semialdehyde dehydrogenase (1.2.1.27)
Propionyl-CoA carboxylase (EC 6.4.1.3)
Methylmalonyl-CoA mutase (EC 5.4.99.2)
Isoleucine (to succinyl CoA and acetyl CoA)
Enoyl-CoA hydratase (EC 4.2.1.17)
Methyl-hydroxybutyryl CoA dehydrogenase (EC 1.1.1.178)
3-ketoacyl-CoA thiolase (EC 2.3.1.16)
Propionyl-CoA carboxylase (EC 6.4.1.3
Methylmalonyl-CoA mutase (EC 5.4.99.2)
Leucine (to acetyl CoA)
Methylcrotonoyl-CoA carboxylase (EC 6.4.1.4)
Methylglutaconyl-CoA hydratase (EC 4.2.1.18)
Hydroxymethylglutaryl-CoA lyase (EC 4.1.3.4)
Methionine (to cysteine and succinyl-CoA):
Methionine adenosyltransferase (EC 2.5.1.6)
Methyltransferase (eg. 2.1.1.37)
Adenosylhomocysteinase (EC 3.3.1.1)
Cystathionine beta-synthase (EC 4.2.1.22)
Cystathionine gamma-lyase (EC 4.4.1.1)
2-oxobutanoate dehydrogenase (EC 1.2.4.4)
Propionyl-CoA carboxylase (EC 6.4.1.3)
Methylmalonyl-CoA mutase (EC 5.4.99.2)
Lysine (to glutaryl-CoA):
Saccharopine dehydrogenase (EC 1.5.1.8)
Saccharopine reductase (EC 1.5.1.10)
Aminoadipate-semialdehyde dehydrogenase (EC 1.2.1.31)
2-aminoadipate transaminase (EC2.6.1.39)
2-oxoglutarate dehydrogenase (EC 1.2.4.2)
432
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
Appendix: EC Numbers
Phenylalanine (to tyrosine):
Phenylalanine hydroxylase (EC 1.14.16.1)
Tyrosine (to fumarate and acetoacetate):
Tyrosine transaminase (EC 2.6.1.5)
4-hydroxyphenylpyruvate dioxygenase (EC 1.13.11.27)
Homogentisate 1,2-dioxygenase (EC 1.13.11.5)
Maleylacetoacetate isomerase (EC 5.2.1.2)
Fumarylacetoacetase (EC 3.7.1.2)
Organic Chemistry With a Biological Emphasis (2016 ed.)
Tim Soderberg
433